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# Ex.2.5 Q8 Polynomials Solution - NCERT Maths Class 9

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## Question

Factorise each of the following:

(i) \begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}\end{align}

(ii) \begin{align}8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\end{align}

(iii) \begin{align}27-125 a^{3}-135 a+225 a^{2}\end{align}

(iv) \begin{align}64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\end{align}

(v) \begin{align}27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\end{align}

Video Solution
Polynomials
Ex 2.5 | Question 8

## Text Solution

Reasoning:

Identities:

\begin{align}&(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}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}\end{align}

This can be re-written as:

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

Which is of the form:

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

Hence,

\begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}=(2 a+b)^{3}\end{align}

(ii) \begin{align}8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\end{align}

This can be re-written as:

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

Which is of the form:

\begin{align}x^{3}-y^{3}-3 x^{2} y+3 x y^{2}=(x-y)^{3}\end{align}

Hence,

\begin{align}8 a^{3}\!-\!b^{3}\!-\!12 a^{2} b\!+\!6 a b^{2}\!=\!(2 a\!-\!b)^{3}\end{align}

(iii)\begin{align}27-125 a^{3}-135 a+225 a^{2}\end{align}

This can be re-witten as:

\begin{align}&(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2} \\&(3)^{3}-(5 a)^{3}-3(3)(5 a)(3-5 a) & \end{align}

Which is of the form:

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

Hence,

\begin{align}27-125 a^{3}-135 a+225 a^{2}=(3-5 a)^{3}\end{align}

(iv) \begin{align} 64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\end{align}

This can be re-written as:

\begin{align}&(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2} \\ &(4 a)^{3}-(3 b)^{3}-3(4 a)(3 b)(4 a-3 b) \end{align}

Which is of the form:

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

Hence,

\begin{align}64 a^{3}\!-\!27 b^{3}\!-\!144 a^{2} b\!+\!108 a b^{2}\!=\!(4 a\!-\!3 b)^{3}\end{align}

(v) \begin{align}\;\;27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\end{align}

This can be re-written as:

\begin{align}&(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)^{2} \frac{1}{6}+3(3 p)\left(\frac{1}{6}\right)^{2} \\ &(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p) \frac{1}{6}\left(3 p-\frac{1}{6}\right)\end{align}

Which is of the form:

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

Hence,

\begin{align}27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p=\left(3 p-\frac{1}{6}\right)^{3}\end{align}

Video Solution
Polynomials
Ex 2.5 | Question 8

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