Ex.2.5 Q6 Polynomials Solution - NCERT Maths Class 9

Go back to  'Ex.2.5'

Question

 Write the following cubes in expanded form:

(i) \(\begin{align}(2 x+1)^{3} \end{align}\)

(ii) \(\begin{align}(2 a-3 b)^{3}\end{align}\)

(iii) \(\begin{align}\left(\frac{3}{2} x+1\right)^{3} \end{align}\)

(iv) \(\begin{align}\left(x-\frac{2}{3} y\right)^{3}\end{align}\)

   

 Video Solution
Polynomials
Ex 2.5 | Question 6

Text Solution

  

Reasoning:

\(\begin{align} \rm\bf { Identities: } &(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y) \\ &(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y) \end{align}\)

Steps:

(i) \(\begin{align}(2 x+1)^{3}\end{align}\)

Identity: \(\begin{align}(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\end{align}\)

Here \(\begin{align}x=2 x\;,\; y=1\end{align}\)

\[\begin{align}(2 x+1)^{3} &=(2 x)^{3}+(1)^{3}+3(2 x)(1)(2 x+1) \\ &=8 x^{3}+1+6 x(2 x+1) \\ &=8 x^{3}+1+12 x^{2}+6 x \\ &=8 x^{3}+12 x^{2}+6 x+1 \end{align}\]

(ii) \(\begin{align}\left(2 a-3 b)^{3}\right.\end{align}\)

Identity: \(\begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}\)

Here \(x = 2a, y = 3b\)

\[\begin{align}(2 a-3 b)^{3} &=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)(2 a-3 b) \\ &=8 a^{3}-27 b^{3}-18 a b(2 a-3 b) \\ &=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2} \\ &=8 a^{3}-36 a^{2} b+54 a b^{2}-27 b^{3} \end{align}\]

(iii) \(\begin{align}\left[\frac{3}{2} x+1\right]^{3} \end{align}\)

Identity: \(\begin{align}(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\end{align}\)

Here \(\begin{align}x=\frac{3}{2}, y=1\end{align}\)

\[\begin{align}\left(\frac{3}{2} x+1\right)^{3} &=\left(\frac{3}{2}x\right)^{3}+(1)^{3}+3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x+1\right) \\ &=\frac{27}{8} x^{3}+1+\frac{9}{2} x+\left(\frac{3}{2} x+1\right) \\ &=\frac{27}{8} x^{3}+1+\frac{27}{8} x^{2}+\frac{9}{2} x \\ &=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1 \end{align}\]

(iv) \(\begin{align}\left(x-\frac{2}{3} y\right)^{3} \end{align}\)

Identity: \(\begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}\)

Here \(\begin{align}x=x, y=\frac{2}{3} y\end{align}\)

\[\begin{align}\left(x-\frac{2}{3} y\right)^{3} &=x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right) \\ &=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right) \\ &=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2} \\ &=x^{3}-2 x^{2} y+\frac{4}{3} x y^{2}-\frac{8}{27} y^{3} \end{align}\]