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:

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

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