Ex.2.5 Q11 Polynomials Solution - NCERT Maths Class 9

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Question

Factorise: \(\begin{align}27 x^{3}+y^{3}+z^{3}-9 x y z\end{align}\)

 

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Polynomials
Ex 2.5 | Question 11

Text Solution

  

Reasoning:

Identity: \(x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)\)

Steps:

The above expression can be written as: \((3 x)^{3}+(y)^{3}+(z)^{2}-3(3 x)(y)(z)\)

\(\begin{align}\text{By using the identity: }x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)\end{align}\)

\[\begin{align}\text{We can write}\quad &{(3 x)^{3}+(y)^{3}+(z)^{2}-3(3 x)(y)(z)} \\ &={(3 x+y+z)\left[(3 x)^{2}+(y)^{2}+(z)^{2}-(3 x)(y)-y z-(z)(3 x)\right]}\end{align}\]

\(\begin{align}\text{Hence }27 x^{3}+y^{3}+z^{3}-9 x y z=(3 x+y+z)\left(9 x^{2}+y^{2}+z^{2}-3 x y-y z-3 z x\right)\end{align}\)