Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Question 1. Use suitable identities to find the following products:
(i) $(-3x+4)^2$
Solution:
We know the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Rewrite the given expression as
$$ (-3x+4)^2=(4-3x)^2. $$Here,
$$ a=4 \qquad\text{and}\qquad b=3x. $$Substitute these values into the identity.
$$ \begin{aligned} (4-3x)^2 &=4^2-2(4)(3x)+(3x)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} 4^2&=16,\\[4pt] 2(4)(3x)&=24x,\\[4pt] (3x)^2&=9x^2. \end{aligned} $$Therefore,
$$ \begin{aligned} (-3x+4)^2 &=16-24x+9x^2\\ &=9x^2-24x+16. \end{aligned} $$Answer:
$$ \boxed{(-3x+4)^2=9x^2-24x+16} $$(ii) $(2s+7)(2s-7)$
Solution:
We know the identity
$$ (a+b)(a-b)=a^2-b^2. $$Here,
$$ a=2s \qquad\text{and}\qquad b=7. $$Substitute these values into the identity.
$$ \begin{aligned} (2s+7)(2s-7) &=(2s)^2-7^2. \end{aligned} $$Now simplify.
$$ \begin{aligned} (2s)^2&=4s^2,\\[4pt] 7^2&=49. \end{aligned} $$Therefore,
$$ \begin{aligned} (2s+7)(2s-7) &=4s^2-49. \end{aligned} $$Answer:
$$ \boxed{(2s+7)(2s-7)=4s^2-49} $$(iii) $\left(p^2+\dfrac12\right)\left(p^2-\dfrac12\right)$
Solution:
We know the identity
$$ (a+b)(a-b)=a^2-b^2. $$Here,
$$ a=p^2 \qquad\text{and}\qquad b=\frac12. $$Substitute these values into the identity.
$$ \begin{aligned} \left(p^2+\frac12\right)\left(p^2-\frac12\right) &=(p^2)^2-\left(\frac12\right)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} (p^2)^2&=p^4,\\[4pt] \left(\frac12\right)^2&=\frac14. \end{aligned} $$Therefore,
$$ \begin{aligned} \left(p^2+\frac12\right)\left(p^2-\frac12\right) &=p^4-\frac14. \end{aligned} $$Answer:
$$ \boxed{\left(p^2+\frac12\right)\left(p^2-\frac12\right)=p^4-\frac14} $$(iv) $(2n+7)(2n-7)$
Solution:
We know the identity
$$ (a+b)(a-b)=a^2-b^2. $$Here,
$$ a=2n \qquad\text{and}\qquad b=7. $$Substitute these values into the identity.
$$ \begin{aligned} (2n+7)(2n-7) &=(2n)^2-7^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} (2n)^2&=4n^2,\\[4pt] 7^2&=49. \end{aligned} $$Therefore,
$$ \begin{aligned} (2n+7)(2n-7) &=4n^2-49. \end{aligned} $$Answer:
$$ \boxed{(2n+7)(2n-7)=4n^2-49} $$(v) $(s-2t)(s^2+2st+4t^2)$
Solution:
We know the identity
$$ (a-b)(a^2+ab+b^2)=a^3-b^3. $$Here,
$$ a=s \qquad\text{and}\qquad b=2t. $$Substitute these values into the identity.
$$ \begin{aligned} (s-2t)(s^2+2st+4t^2) &=s^3-(2t)^3. \end{aligned} $$Now simplify.
$$ \begin{aligned} (2t)^3 &=8t^3. \end{aligned} $$Therefore,
$$ \begin{aligned} (s-2t)(s^2+2st+4t^2) &=s^3-8t^3. \end{aligned} $$Answer:
$$ \boxed{(s-2t)(s^2+2st+4t^2)=s^3-8t^3} $$(vi) $\left(\dfrac{1}{2r}-4r\right)^2$
Solution:
We know the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Here,
$$ a=\frac{1}{2r} \qquad\text{and}\qquad b=4r. $$Substitute these values into the identity.
$$ \begin{aligned} \left(\frac{1}{2r}-4r\right)^2 &=\left(\frac{1}{2r}\right)^2 -2\left(\frac{1}{2r}\right)(4r) +(4r)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} \left(\frac{1}{2r}\right)^2 &=\frac{1}{4r^2},\\[6pt] 2\left(\frac{1}{2r}\right)(4r) &=4,\\[6pt] (4r)^2 &=16r^2. \end{aligned} $$Substituting these values, we get
$$ \begin{aligned} \left(\frac{1}{2r}-4r\right)^2 &=\frac{1}{4r^2}-4+16r^2. \end{aligned} $$Writing the terms in descending order,
$$ \boxed{ \left(\frac{1}{2r}-4r\right)^2 = 16r^2-4+\frac{1}{4r^2} } $$(vii) $(-3m+4k-l)^2$
Solution:
We know the identity
$$ (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca. $$Rewrite the given expression as
$$ (-3m+4k-l)^2=(4k-3m-l)^2. $$Here,
$$ a=4k,\qquad b=-3m,\qquad c=-l. $$Substitute these values into the identity.
$$ \begin{aligned} (4k-3m-l)^2 &=(4k)^2+(-3m)^2+(-l)^2\\ &\quad+2(4k)(-3m)\\ &\quad+2(-3m)(-l)\\ &\quad+2(4k)(-l). \end{aligned} $$Now simplify each term.
$$ \begin{aligned} (4k)^2&=16k^2,\\[4pt] (-3m)^2&=9m^2,\\[4pt] (-l)^2&=l^2,\\[4pt] 2(4k)(-3m)&=-24km,\\[4pt] 2(-3m)(-l)&=6ml,\\[4pt] 2(4k)(-l)&=-8kl. \end{aligned} $$Therefore,
$$ \begin{aligned} (-3m+4k-l)^2 &=16k^2+9m^2+l^2\\ &\quad-24km-8kl+6ml. \end{aligned} $$Answer:
$$ \boxed{ (-3m+4k-l)^2 = 16k^2+9m^2+l^2-24km-8kl+6ml } $$(viii) $\left(x-\dfrac13y\right)^3$
Solution:
We know the identity
$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3. $$Here,
$$ a=x \qquad\text{and}\qquad b=\frac13y. $$Substitute these values into the identity.
$$ \begin{aligned} \left(x-\frac13y\right)^3 &=x^3 -3x^2\left(\frac13y\right) +3x\left(\frac13y\right)^2 -\left(\frac13y\right)^3. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} 3x^2\left(\frac13y\right) &=x^2y,\\[6pt] 3x\left(\frac13y\right)^2 &=3x\left(\frac19y^2\right) =\frac13xy^2,\\[6pt] \left(\frac13y\right)^3 &=\frac1{27}y^3. \end{aligned} $$Therefore,
$$ \begin{aligned} \left(x-\frac13y\right)^3 &=x^3-x^2y+\frac13xy^2-\frac1{27}y^3. \end{aligned} $$Answer:
$$ \boxed{ \left(x-\frac13y\right)^3 = x^3-x^2y+\frac13xy^2-\frac1{27}y^3 } $$(ix) $\left(\dfrac72k-\dfrac23m\right)^3$
Solution:
We know the identity
$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3. $$Here,
$$ a=\frac72k \qquad\text{and}\qquad b=\frac23m. $$Substitute these values into the identity.
$$ \begin{aligned} \left(\frac72k-\frac23m\right)^3 &=\left(\frac72k\right)^3 -3\left(\frac72k\right)^2\left(\frac23m\right)\\ &\quad+3\left(\frac72k\right)\left(\frac23m\right)^2 -\left(\frac23m\right)^3. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} \left(\frac72k\right)^3 &=\frac{343}{8}k^3,\\[6pt] 3\left(\frac72k\right)^2\left(\frac23m\right) &=3\left(\frac{49}{4}k^2\right)\left(\frac23m\right)\\ &=\frac{49}{2}k^2m,\\[6pt] 3\left(\frac72k\right)\left(\frac23m\right)^2 &=3\left(\frac72k\right)\left(\frac49m^2\right)\\ &=\frac{14}{3}km^2,\\[6pt] \left(\frac23m\right)^3 &=\frac{8}{27}m^3. \end{aligned} $$Therefore,
$$ \begin{aligned} \left(\frac72k-\frac23m\right)^3 &=\frac{343}{8}k^3 -\frac{49}{2}k^2m +\frac{14}{3}km^2 -\frac{8}{27}m^3. \end{aligned} $$Answer:
$$ \boxed{ \left(\frac72k-\frac23m\right)^3 = \frac{343}{8}k^3 -\frac{49}{2}k^2m +\frac{14}{3}km^2 -\frac{8}{27}m^3 } $$Question 2. Find the values using suitable identities:
(i) $17 \times 21$
Solution:
Observe that
$$ 17=19-2 \qquad\text{and}\qquad 21=19+2. $$We use the identity
$$ (a-b)(a+b)=a^2-b^2. $$Here,
$$ a=19 \qquad\text{and}\qquad b=2. $$Using the identity,
$$ \begin{aligned} 17\times21 &=(19-2)(19+2)\\ &=19^2-2^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 19^2&=361,\\ 2^2&=4. \end{aligned} $$Therefore,
$$ \begin{aligned} 17\times21 &=361-4\\ &=357. \end{aligned} $$Answer:
$$ \boxed{17\times21=357} $$(ii) $104 \times 96$
Solution:
Observe that
$$ 104=100+4 \qquad\text{and}\qquad 96=100-4. $$We use the identity
$$ (a+b)(a-b)=a^2-b^2. $$Here,
$$ a=100 \qquad\text{and}\qquad b=4. $$Using the identity,
$$ \begin{aligned} 104\times96 &=(100+4)(100-4)\\ &=100^2-4^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 100^2&=10000,\\ 4^2&=16. \end{aligned} $$Therefore,
$$ \begin{aligned} 104\times96 &=10000-16\\ &=9984. \end{aligned} $$Answer:
$$ \boxed{104\times96=9984} $$(iii) $24 \times 16$
Solution:
Observe that
$$ 24=20+4 \qquad\text{and}\qquad 16=20-4. $$We use the identity
$$ (a+b)(a-b)=a^2-b^2. $$Here,
$$ a=20 \qquad\text{and}\qquad b=4. $$Using the identity,
$$ \begin{aligned} 24\times16 &=(20+4)(20-4)\\ &=20^2-4^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 20^2&=400,\\ 4^2&=16. \end{aligned} $$Therefore,
$$ \begin{aligned} 24\times16 &=400-16\\ &=384. \end{aligned} $$Answer:
$$ \boxed{24\times16=384} $$(iv) $147^3$
Solution:
Since $147$ is close to $150$, we write
$$ 147=150-3. $$We use the identity
$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3. $$Here,
$$ a=150 \qquad\text{and}\qquad b=3. $$Using the identity,
$$ \begin{aligned} 147^3 &=(150-3)^3\\ &=150^3-3(150)^2(3)\\ &\quad+3(150)(3)^2-3^3. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 150^3&=3375000,\\[4pt] 3(150)^2(3)&=202500,\\[4pt] 3(150)(3)^2&=4050,\\[4pt] 3^3&=27. \end{aligned} $$Substitute these values.
$$ \begin{aligned} 147^3 &=3375000-202500+4050-27\\ &=3176523. \end{aligned} $$Answer:
$$ \boxed{147^3=3176523} $$(v) $199^3$
Solution:
Since $199$ is close to $200$, we write
$$ 199=200-1. $$We use the identity
$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3. $$Here,
$$ a=200 \qquad\text{and}\qquad b=1. $$Using the identity,
$$ \begin{aligned} 199^3 &=(200-1)^3\\ &=200^3-3(200)^2(1)\\ &\quad+3(200)(1)^2-1^3. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 200^3&=8000000,\\[4pt] 3(200)^2(1)&=120000,\\[4pt] 3(200)(1)^2&=600,\\[4pt] 1^3&=1. \end{aligned} $$Substitute these values.
$$ \begin{aligned} 199^3 &=8000000-120000+600-1\\ &=7880599. \end{aligned} $$Answer:
$$ \boxed{199^3=7880599} $$(vi) $127^3$
Solution:
Since $127$ is close to $130$, we write
$$ 127=130-3. $$We use the identity
$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3. $$Here,
$$ a=130 \qquad\text{and}\qquad b=3. $$Using the identity,
$$ \begin{aligned} 127^3 &=(130-3)^3\\ &=130^3-3(130)^2(3)\\ &\quad+3(130)(3)^2-3^3. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 130^3&=2197000,\\[4pt] 3(130)^2(3)&=152100,\\[4pt] 3(130)(3)^2&=3510,\\[4pt] 3^3&=27. \end{aligned} $$Substitute these values.
$$ \begin{aligned} 127^3 &=2197000-152100+3510-27\\ &=2048383. \end{aligned} $$Answer:
$$ \boxed{127^3=2048383} $$(vii) $(-107)^3$
Solution:
Write
$$ -107=-(100+7). $$First, find $(100+7)^3$ using the identity
$$ (a+b)^3=a^3+3a^2b+3ab^2+b^3. $$Here,
$$ a=100 \qquad\text{and}\qquad b=7. $$Using the identity,
$$ \begin{aligned} (100+7)^3 &=100^3+3(100)^2(7)\\ &\quad+3(100)(7)^2+7^3. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 100^3&=1000000,\\[4pt] 3(100)^2(7)&=210000,\\[4pt] 3(100)(7)^2&=14700,\\[4pt] 7^3&=343. \end{aligned} $$Therefore,
$$ \begin{aligned} (107)^3 &=1000000+210000+14700+343\\ &=1225043. \end{aligned} $$Since the cube of a negative number is negative,
$$ \boxed{(-107)^3=-1225043.} $$(viii) $(-299)^3$
Solution:
Write
$$ -299=-(300-1). $$First, find $(300-1)^3$ using the identity
$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3. $$Here,
$$ a=300 \qquad\text{and}\qquad b=1. $$Using the identity,
$$ \begin{aligned} (300-1)^3 &=300^3-3(300)^2(1)\\ &\quad+3(300)(1)^2-1^3. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 300^3&=27000000,\\[4pt] 3(300)^2(1)&=270000,\\[4pt] 3(300)(1)^2&=900,\\[4pt] 1^3&=1. \end{aligned} $$Substitute these values.
$$ \begin{aligned} 299^3 &=27000000-270000+900-1\\ &=26730999. \end{aligned} $$Since the cube of a negative number is negative,
$$ \boxed{(-299)^3=-26730999.} $$Question 3. Factor the following algebraic expressions:
(i) $4y^2+1+\dfrac{1}{16y^2}$
Solution:
We know the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Write each term as a perfect square.
$$ 4y^2=(2y)^2, $$ $$ 1=2(2y)\left(\frac{1}{4y}\right), $$ $$ \frac{1}{16y^2}=\left(\frac{1}{4y}\right)^2. $$Hence, the given expression is of the form
$$ a^2+2ab+b^2, $$where
$$ a=2y,\qquad b=\frac{1}{4y}. $$Therefore,
$$ \boxed{ 4y^2+1+\frac{1}{16y^2} = \left(2y+\frac{1}{4y}\right)^2 } $$(ii) $9m^2-\dfrac{1}{25n^2}$
Solution:
We know the identity
$$ a^2-b^2=(a-b)(a+b). $$Write each term as a perfect square.
$$ 9m^2=(3m)^2, $$ $$ \frac{1}{25n^2}=\left(\frac{1}{5n}\right)^2. $$Using the identity,
$$ \begin{aligned} 9m^2-\frac{1}{25n^2} &=(3m)^2-\left(\frac{1}{5n}\right)^2\\ &=\left(3m-\frac{1}{5n}\right) \left(3m+\frac{1}{5n}\right). \end{aligned} $$Answer:
$$ \boxed{ 9m^2-\frac{1}{25n^2} = \left(3m-\frac{1}{5n}\right) \left(3m+\frac{1}{5n}\right) } $$(iii) $27b^3-\dfrac{1}{64b^3}$
Solution:
We know the identity
$$ a^3-b^3=(a-b)(a^2+ab+b^2). $$Write each term as a perfect cube.
$$ 27b^3=(3b)^3, $$ $$ \frac{1}{64b^3} = \left(\frac{1}{4b}\right)^3. $$Here,
$$ a=3b, \qquad b=\frac{1}{4b}. $$Using the identity,
$$ \begin{aligned} 27b^3-\frac{1}{64b^3} &= \left(3b-\frac{1}{4b}\right) \left[ (3b)^2 +(3b)\left(\frac{1}{4b}\right) +\left(\frac{1}{4b}\right)^2 \right]. \end{aligned} $$Now simplify the terms inside the bracket.
$$ (3b)^2=9b^2, $$ $$ (3b)\left(\frac{1}{4b}\right)=\frac34, $$ $$ \left(\frac{1}{4b}\right)^2=\frac{1}{16b^2}. $$Hence,
$$ \boxed{ 27b^3-\frac{1}{64b^3} = \left(3b-\frac{1}{4b}\right) \left( 9b^2+\frac34+\frac{1}{16b^2} \right) } $$(iv) $x^2+\dfrac{5x}{6}+\dfrac16$
Solution:
We have to factorise the given quadratic expression.
$$ x^2+\frac{5x}{6}+\frac16. $$Multiply the constant term and the coefficient of $x^2$.
$$ 1\times\frac16=\frac16. $$We need two numbers whose product is
$$ \frac16 $$and whose sum is
$$ \frac56. $$The required numbers are
$$ \frac12 \qquad\text{and}\qquad \frac13, $$because
$$ \frac12+\frac13=\frac56, $$and
$$ \frac12\times\frac13=\frac16. $$Split the middle term.
$$ \begin{aligned} x^2+\frac{5x}{6}+\frac16 &=x^2+\frac12x+\frac13x+\frac16. \end{aligned} $$Group the terms.
$$ \begin{aligned} &=x\left(x+\frac12\right) +\frac13\left(x+\frac12\right). \end{aligned} $$Take the common factor.
$$ \begin{aligned} &=\left(x+\frac13\right) \left(x+\frac12\right). \end{aligned} $$Answer:
$$ \boxed{ \left(x+\frac13\right) \left(x+\frac12\right) } $$(v) $27u^3-\dfrac{1}{125}-\dfrac{27u^2}{5}+\dfrac{9u}{25}$
Solution:
Rearrange the terms.
$$ 27u^3-\frac{27u^2}{5} +\frac{9u}{25} -\frac{1}{125}. $$We know the identity
$$ (a-b)^3 = a^3-3a^2b+3ab^2-b^3. $$Write each term in the required form.
$$ 27u^3=(3u)^3, $$ $$ \frac{1}{125} =\left(\frac15\right)^3. $$Now check the middle terms.
$$ 3(3u)^2\left(\frac15\right) = \frac{27u^2}{5}, $$ $$ 3(3u)\left(\frac15\right)^2 = \frac{9u}{25}. $$Hence, the given expression is of the form
$$ a^3-3a^2b+3ab^2-b^3, $$where
$$ a=3u, \qquad b=\frac15. $$Therefore,
$$ \boxed{ 27u^3-\frac{1}{125} -\frac{27u^2}{5} +\frac{9u}{25} = \left(3u-\frac15\right)^3 } $$(vi) $64y^3+\dfrac{1}{125}z^3$
Solution:
We know the identity
$$ a^3+b^3 = (a+b)(a^2-ab+b^2). $$Write each term as a perfect cube.
$$ 64y^3=(4y)^3, $$ $$ \frac{1}{125}z^3 = \left(\frac{z}{5}\right)^3. $$Here,
$$ a=4y, \qquad b=\frac{z}{5}. $$Using the identity,
$$ \begin{aligned} 64y^3+\frac{1}{125}z^3 &= \left(4y+\frac{z}{5}\right) \left[ (4y)^2 -(4y)\left(\frac{z}{5}\right) +\left(\frac{z}{5}\right)^2 \right]. \end{aligned} $$Now simplify the terms inside the bracket.
$$ (4y)^2=16y^2, $$ $$ (4y)\left(\frac{z}{5}\right) =\frac{4yz}{5}, $$ $$ \left(\frac{z}{5}\right)^2 =\frac{z^2}{25}. $$Hence,
$$ \boxed{ 64y^3+\frac{1}{125}z^3 = \left(4y+\frac{z}{5}\right) \left( 16y^2-\frac{4yz}{5}+\frac{z^2}{25} \right) } $$(vii) $p^3+27q^3+r^3-9pqr$
Solution:
We know the identity
$$ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca). $$Compare the given expression with the identity.
Here,
$$ a=p, \qquad b=3q, \qquad c=r. $$Since
$$ (3q)^3=27q^3, $$and
$$ 3(p)(3q)(r)=9pqr, $$the given expression is exactly of the form
$$ a^3+b^3+c^3-3abc. $$Applying the identity,
$$ \begin{aligned} p^3+27q^3+r^3-9pqr &=(p+3q+r)\\ &\quad\left(p^2+9q^2+r^2-3pq-3qr-pr\right). \end{aligned} $$Answer:
$$ \boxed{ (p+3q+r) \left(p^2+9q^2+r^2-3pq-3qr-pr\right) } $$(viii) $9m^2-12m+4$
Solution:
We know the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Write each term as a perfect square.
$$ 9m^2=(3m)^2, $$ $$ 4=2^2. $$Now check the middle term.
$$ 2(3m)(2)=12m. $$Hence, the given expression is of the form
$$ a^2-2ab+b^2, $$where
$$ a=3m, \qquad b=2. $$Therefore,
$$ \boxed{ 9m^2-12m+4=(3m-2)^2 } $$Question 4. Simplify the following:
Note: Assume that the denominators are not equal to $0$.
(i) $\dfrac{4x^2+4x+1}{4x^2-1}$
Solution:
The given expression is
$$ \frac{4x^2+4x+1}{4x^2-1}. $$Step 1: Factorise the numerator.
Using the identity
$$ a^2+2ab+b^2=(a+b)^2, $$where
$$ a=2x,\qquad b=1, $$we get
$$ \begin{aligned} 4x^2+4x+1 &=(2x+1)^2. \end{aligned} $$Step 2: Factorise the denominator.
Using the identity
$$ a^2-b^2=(a-b)(a+b), $$where
$$ a=2x,\qquad b=1, $$we obtain
$$ \begin{aligned} 4x^2-1 &=(2x)^2-1^2\\ &=(2x-1)(2x+1). \end{aligned} $$Step 3: Cancel the common factor.
$$ \begin{aligned} \frac{(2x+1)^2} {(2x-1)(2x+1)} &=\frac{2x+1}{2x-1}. \end{aligned} $$Answer:
$$ \boxed{ \frac{2x+1}{2x-1} } $$(ii) $\dfrac{9(3a^3-24b^3)}{9a^2-36b^2}$
Solution:
The given expression is
$$ \frac{9(3a^3-24b^3)} {9a^2-36b^2}. $$Step 1: Factorise the numerator.
Take $3$ common from the bracket.
$$ \begin{aligned} 3a^3-24b^3 &=3(a^3-8b^3). \end{aligned} $$Therefore,
$$ \begin{aligned} 9(3a^3-24b^3) &=27(a^3-8b^3). \end{aligned} $$Now use the identity
$$ a^3-b^3=(a-b)(a^2+ab+b^2). $$Here,
$$ a=a,\qquad b=2b. $$Hence,
$$ \begin{aligned} a^3-8b^3 &=(a-2b)(a^2+2ab+4b^2). \end{aligned} $$Thus, the numerator becomes
$$ 27(a-2b)(a^2+2ab+4b^2). $$Step 2: Factorise the denominator.
Take $9$ common.
$$ \begin{aligned} 9a^2-36b^2 &=9(a^2-4b^2). \end{aligned} $$Using the identity
$$ a^2-b^2=(a-b)(a+b), $$we get
$$ \begin{aligned} 9(a^2-4b^2) &=9(a-2b)(a+2b). \end{aligned} $$Step 3: Cancel the common factors.
$$ \begin{aligned} \frac{27(a-2b)(a^2+2ab+4b^2)} {9(a-2b)(a+2b)} &= \frac{3(a^2+2ab+4b^2)} {a+2b}. \end{aligned} $$Answer:
$$ \boxed{ \frac{3(a^2+2ab+4b^2)} {a+2b} } $$(iii) $\dfrac{s^3+125t^3}{s^2-2st-35t^2}$
Solution:
The given expression is
$$ \frac{s^3+125t^3} {s^2-2st-35t^2}. $$Step 1: Factorise the numerator.
Using the identity
$$ a^3+b^3=(a+b)(a^2-ab+b^2), $$where
$$ a=s,\qquad b=5t, $$we get
$$ \begin{aligned} s^3+125t^3 &=(s+5t)(s^2-5st+25t^2). \end{aligned} $$Step 2: Factorise the denominator.
We need two numbers whose product is
$$ -35 $$and whose sum is
$$ -2. $$The required numbers are
$$ 5 \qquad\text{and}\qquad -7. $$Split the middle term.
$$ \begin{aligned} s^2-2st-35t^2 &=s^2+5st-7st-35t^2\\ &=s(s+5t)-7t(s+5t)\\ &=(s+5t)(s-7t). \end{aligned} $$Step 3: Cancel the common factor.
$$ \begin{aligned} \frac{(s+5t)(s^2-5st+25t^2)} {(s+5t)(s-7t)} &= \frac{s^2-5st+25t^2} {s-7t}. \end{aligned} $$Answer:
$$ \boxed{ \frac{s^2-5st+25t^2} {s-7t} } $$Question 5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) $25a^2-30ab+9b^2$
Solution:
The area of a rectangle is given by
$$ \text{Area}=\text{Length}\times\text{Breadth}. $$Therefore, we factorise the given expression.
The given expression is
$$ 25a^2-30ab+9b^2. $$We know the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Write each term as a perfect square.
$$ 25a^2=(5a)^2, $$ $$ 9b^2=(3b)^2. $$Now check the middle term.
$$ 2(5a)(3b)=30ab. $$Hence, the given expression is of the form
$$ a^2-2ab+b^2, $$where
$$ a=5a, \qquad b=3b. $$Therefore,
$$ \begin{aligned} 25a^2-30ab+9b^2 &=(5a-3b)^2\\ &=(5a-3b)(5a-3b). \end{aligned} $$Hence, one possible pair of dimensions is
$$ \boxed{ \text{Length}=5a-3b,\qquad \text{Breadth}=5a-3b. } $$(ii) $36s^2-49t^2$
Solution:
The area of a rectangle is
$$ \text{Area}=\text{Length}\times\text{Breadth}. $$Therefore, factorise the given expression.
The given expression is
$$ 36s^2-49t^2. $$We know the identity
$$ a^2-b^2=(a-b)(a+b). $$Write each term as a perfect square.
$$ 36s^2=(6s)^2, $$ $$ 49t^2=(7t)^2. $$Applying the identity,
$$ \begin{aligned} 36s^2-49t^2 &=(6s-7t)(6s+7t). \end{aligned} $$Hence, one possible pair of dimensions is
$$ \boxed{ \text{Length}=6s-7t,\qquad \text{Breadth}=6s+7t. } $$Interchanging the length and breadth also gives a correct answer.
Question 6. Find possible expressions for the length, breadth, and height of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) $6a^2-24b^2$
Solution:
The volume of a cuboid is given by
$$ \text{Volume}=\text{Length}\times\text{Breadth}\times\text{Height}. $$Therefore, we factorise the given expression.
The given expression is
$$ 6a^2-24b^2. $$Step 1: Take the common factor.
$$ \begin{aligned} 6a^2-24b^2 &=6(a^2-4b^2). \end{aligned} $$Step 2: Factorise the remaining expression.
Using the identity
$$ a^2-b^2=(a-b)(a+b), $$we get
$$ \begin{aligned} a^2-4b^2 &=a^2-(2b)^2\\ &=(a-2b)(a+2b). \end{aligned} $$Therefore,
$$ \begin{aligned} 6a^2-24b^2 &=6(a-2b)(a+2b). \end{aligned} $$Hence, one possible set of dimensions is
$$ \boxed{ \text{Length}=6,\qquad \text{Breadth}=a-2b,\qquad \text{Height}=a+2b. } $$Any other arrangement of these three factors is also correct.
(ii) $3ps^2-15ps+12p$
Solution:
The volume of a cuboid is
$$ \text{Volume}=\text{Length}\times\text{Breadth}\times\text{Height}. $$Therefore, factorise the given expression.
The given expression is
$$ 3ps^2-15ps+12p. $$Step 1: Take the common factor.
$$ \begin{aligned} 3ps^2-15ps+12p &=3p(s^2-5s+4). \end{aligned} $$Step 2: Factorise the quadratic expression.
We need two numbers whose product is
$$ 4 $$and whose sum is
$$ -5. $$The required numbers are
$$ -1 \qquad\text{and}\qquad -4. $$Hence,
$$ \begin{aligned} s^2-5s+4 &=(s-1)(s-4). \end{aligned} $$Therefore,
$$ \begin{aligned} 3ps^2-15ps+12p &=3p(s-1)(s-4). \end{aligned} $$Hence, one possible set of dimensions is
$$ \boxed{ \text{Length}=3p,\qquad \text{Breadth}=s-1,\qquad \text{Height}=s-4. } $$Any other arrangement of these three factors is also correct.
Question 7. The village playground is shaped as a square of side 40 metres. A path of width $s$ metres is created around the playground for people to walk. Find an expression for the area of the path in terms of $s$.
Solution:
The playground is a square of side
$$ 40\text{ m}. $$A path of width $s$ metres is constructed around the playground.
Therefore, the side of the outer square becomes
$$ 40+2s. $$Step 1: Find the area of the outer square.
$$ \begin{aligned} \text{Area of outer square} &=(40+2s)^2. \end{aligned} $$Using the identity
$$ (a+b)^2=a^2+2ab+b^2, $$we get
$$ \begin{aligned} (40+2s)^2 &=40^2+2(40)(2s)+(2s)^2\\ &=1600+160s+4s^2. \end{aligned} $$Step 2: Find the area of the playground.
$$ \begin{aligned} \text{Area of playground} &=40^2\\ &=1600. \end{aligned} $$Step 3: Find the area of the path.
$$ \begin{aligned} \text{Area of path} &=\text{Area of outer square}-\text{Area of playground}\\ &=(1600+160s+4s^2)-1600\\ &=160s+4s^2. \end{aligned} $$Taking $4s$ as the common factor,
$$ \begin{aligned} 160s+4s^2 &=4s(40+s). \end{aligned} $$Answer:
$$ \boxed{\text{Area of the path}=160s+4s^2=4s(40+s)\text{ square metres}.} $$Question 8. If a number plus its reciprocal equals $\dfrac{10}{3}$, find the number.
Solution:
Let the required number be
$$ x. $$According to the question,
$$ x+\frac1x=\frac{10}{3}. $$Step 1: Remove the denominator.
Multiply both sides by
$$ 3x. $$ $$ \begin{aligned} 3x\left(x+\frac1x\right) &=3x\left(\frac{10}{3}\right). \end{aligned} $$Simplifying,
$$ 3x^2+3=10x. $$Bring all the terms to one side.
$$ 3x^2-10x+3=0. $$Step 2: Factorise the quadratic equation.
We need two numbers whose product is
$$ 3\times3=9 $$and whose sum is
$$ -10. $$The required numbers are
$$ -9 \qquad\text{and}\qquad -1. $$Split the middle term.
$$ \begin{aligned} 3x^2-10x+3 &=3x^2-9x-x+3\\ &=3x(x-3)-1(x-3)\\ &=(3x-1)(x-3). \end{aligned} $$Therefore,
$$ (3x-1)(x-3)=0. $$Hence,
$$ 3x-1=0 \quad\text{or}\quad x-3=0. $$So,
$$ x=\frac13 \quad\text{or}\quad x=3. $$Verification:
$$ 3+\frac13=\frac{10}{3}, $$and
$$ \frac13+3=\frac{10}{3}. $$Both satisfy the given condition.
Answer:
$$ \boxed{x=3\quad\text{or}\quad x=\frac13} $$Question 9. A rectangular pool has area $2x^2+7x+3$ square hastas. If its width is $2x+1$ hastas, find its length.
Solution:
We know that
$$ \text{Area}=\text{Length}\times\text{Width}. $$Given,
$$ \text{Area}=2x^2+7x+3, $$and
$$ \text{Width}=2x+1. $$Therefore,
$$ \text{Length} = \frac{2x^2+7x+3}{2x+1}. $$Step 1: Factorise the numerator.
We need two numbers whose product is
$$ 2\times3=6 $$and whose sum is
$$ 7. $$The required numbers are
$$ 6 \qquad\text{and}\qquad 1. $$Split the middle term.
$$ \begin{aligned} 2x^2+7x+3 &=2x^2+6x+x+3\\ &=2x(x+3)+1(x+3)\\ &=(2x+1)(x+3). \end{aligned} $$Step 2: Divide by the width.
$$ \begin{aligned} \text{Length} &=\frac{(2x+1)(x+3)}{2x+1}\\ &=x+3. \end{aligned} $$Answer:
$$ \boxed{\text{Length}=x+3\text{ hastas}.} $$Question 10. If both $x-2$ and $x-\dfrac12$ are factors of $px^2+5x+r$, show that $p=r$.
Solution:
Let
$$ f(x)=px^2+5x+r. $$Since $x-2$ is a factor of $f(x)$, by the Factor Theorem,
$$ f(2)=0. $$Substitute $x=2$.
$$ \begin{aligned} p(2)^2+5(2)+r&=0\\ 4p+10+r&=0. \end{aligned} $$Therefore,
$$ 4p+r=-10. \qquad\cdots(1) $$Also, since $x-\dfrac12$ is a factor of $f(x)$,
$$ f\left(\frac12\right)=0. $$Substitute $x=\dfrac12$.
$$ \begin{aligned} p\left(\frac12\right)^2 +5\left(\frac12\right) +r &=0. \end{aligned} $$ $$ \begin{aligned} \frac{p}{4} +\frac52 +r &=0. \end{aligned} $$Multiply both sides by $4$.
$$ \begin{aligned} p+10+4r &=0. \end{aligned} $$Therefore,
$$ p+4r=-10. \qquad\cdots(2) $$Subtract equation (2) from equation (1).
$$ \begin{aligned} (4p+r)-(p+4r) &=-10-(-10). \end{aligned} $$ $$ 3p-3r=0. $$Divide both sides by $3$.
$$ p-r=0. $$Hence,
$$ \boxed{p=r.} $$Question 11. If $a+b+c=5$ and $ab+bc+ca=10$, then prove that $a^3+b^3+c^3-3abc=-25$.
Solution:
We know the identity
$$ a^3+b^3+c^3-3abc = (a+b+c) \left( a^2+b^2+c^2-ab-bc-ca \right). $$Given,
$$ a+b+c=5 $$and
$$ ab+bc+ca=10. $$First, find the value of
$$ a^2+b^2+c^2. $$Using the identity
$$ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca), $$we get
$$ \begin{aligned} 5^2 &= a^2+b^2+c^2+2(10). \end{aligned} $$ $$ 25=a^2+b^2+c^2+20. $$Therefore,
$$ a^2+b^2+c^2=5. $$Now substitute the given values into the first identity.
$$ \begin{aligned} a^3+b^3+c^3-3abc &= 5(5-10). \end{aligned} $$ $$ \begin{aligned} &=5(-5)\\ &=-25. \end{aligned} $$Hence,
$$ \boxed{ a^3+b^3+c^3-3abc=-25. } $$Question 12. By factorising the expression, check that $n^3-n$ is always divisible by $6$ for all natural numbers $n$. Give reasons.
Solution:
Consider the given expression
$$ n^3-n. $$Step 1: Take the common factor.
$$ \begin{aligned} n^3-n &=n(n^2-1). \end{aligned} $$Step 2: Factorise further using the identity
$$ a^2-b^2=(a-b)(a+b). $$Since
$$ n^2-1=n^2-1^2, $$we get
$$ \begin{aligned} n^2-1 &=(n-1)(n+1). \end{aligned} $$Therefore,
$$ \begin{aligned} n^3-n &=n(n-1)(n+1). \end{aligned} $$Thus, the expression is the product of three consecutive natural numbers.
Step 3: Show that it is divisible by $2$.
Among any three consecutive natural numbers, at least one number is even.
Therefore,
$$ n(n-1)(n+1) $$is always divisible by
$$ 2. $$Step 4: Show that it is divisible by $3$.
Among any three consecutive natural numbers, one number is always a multiple of $3$.
Therefore,
$$ n(n-1)(n+1) $$is also divisible by
$$ 3. $$Step 5: Conclude the result.
Since the product is divisible by both
$$ 2 \quad\text{and}\quad 3, $$and
$$ 2\times3=6, $$it follows that
$$ n^3-n $$is always divisible by
$$ 6. $$Hence,
$$ \boxed{n^3-n\text{ is divisible by }6\text{ for every natural number }n.} $$Question 13. Find the value of the following:
(i) $x^3+y^3-12xy+64$, when $x+y=-4$
Solution:
Given,
$$ x+y=-4. $$We have to find the value of
$$ x^3+y^3-12xy+64. $$Step 1: Rewrite the constant term.
Since
$$ 64=4^3=(-4)^3(-1), $$write
$$ 64=-(-4)^3. $$Since
$$ x+y=-4, $$we get
$$ 64=-(x+y)^3. $$Therefore,
$$ \begin{aligned} x^3+y^3-12xy+64 &=x^3+y^3-(x+y)^3-12xy. \end{aligned} $$Step 2: Use the identity
$$ (x+y)^3=x^3+y^3+3xy(x+y). $$Substituting this identity,
$$ \begin{aligned} x^3+y^3-12xy+64 &=x^3+y^3-\left[x^3+y^3+3xy(x+y)\right]-12xy. \end{aligned} $$Simplifying,
$$ \begin{aligned} &=-3xy(x+y)-12xy. \end{aligned} $$Now substitute
$$ x+y=-4. $$ $$ \begin{aligned} &=-3xy(-4)-12xy\\ &=12xy-12xy\\ &=0. \end{aligned} $$Answer:
$$ \boxed{0} $$(ii) $x^3-8y^3-36xy-216$, when $x=2y+6$
Solution:
Given,
$$ x=2y+6. $$Therefore,
$$ x-2y=6. $$We have to find
$$ x^3-8y^3-36xy-216. $$Step 1: Rewrite the expression.
Notice that
$$ 8y^3=(2y)^3, $$and
$$ 216=6^3. $$Hence,
$$ x^3-8y^3-36xy-216 = x^3-(2y)^3-6^3-36xy. $$Step 2: Use the identity
$$ a^3+b^3+c^3-3abc = (a+b+c) (a^2+b^2+c^2-ab-bc-ca). $$Take
$$ a=x,\qquad b=-2y,\qquad c=-6. $$Then,
$$ \begin{aligned} a^3+b^3+c^3-3abc &=x^3-(2y)^3-6^3\\ &\quad-3(x)(-2y)(-6). \end{aligned} $$Since
$$ -3(x)(-2y)(-6) =-36xy, $$the given expression becomes
$$ a^3+b^3+c^3-3abc. $$Using the identity,
$$ \begin{aligned} x^3-8y^3-36xy-216 &=(x-2y-6)\\ &\quad\left[x^2+4y^2+36+2xy+12y-6x\right]. \end{aligned} $$Now substitute
$$ x-2y=6. $$Therefore,
$$ x-2y-6=0. $$Hence,
$$ \begin{aligned} x^3-8y^3-36xy-216 &=0\times \left[x^2+4y^2+36+2xy+12y-6x\right]\\ &=0. \end{aligned} $$Answer:
$$ \boxed{0} $$Important identities
Here are some important algebraic identities that are commonly used in mathematics:
Important Algebraic Identities