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Ex.2.5 Q11 Polynomials Solution - NCERT Maths Class 9

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Question

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

 

 Video Solution
Polynomials
Ex 2.5 | Question 11

Text Solution

  

Reasoning:

Identity:

\[\begin{align}&x^{3}+y^{3}+z^{3}-3 x y z\\&\!=\!(x\!+\!y\!+\!z)\!\left(x^{2}\!+\!y^{2}\!+\!z^{2}\!-\!x y\!-\!y z\!-\!z x\right)\end{align}\]

Steps:

The above expression can be written as:

\[(3 x)^{3}+(y)^{3}+(z)^{2}-3(3 x)(y)(z)\]

By using the identity:

\[\begin{align}&x^{3}+y^{3}+z^{3}-3 x y z\\&=(x\!+\!y\!+\!z)\left(x^{2}\!+\!y^{2}\!+\!z^{2}\!-\!x y\!-\!y z\!-\!z x\right)\end{align}\]

We can write:

\[\begin{align} &{(3 x)^{3}+(y)^{3}+(z)^{2}-3(3 x)(y)(z)} \\ &=(3 x+y+z)\!\left[\begin{array}((3 x)^{2}+(y)^{2}+(z)^{2}-\\(3 x)(y)\!-\!y z\!-\!(z)(3 x)\end{array}\right]\end{align}\]

Hence,

\[\begin{align}&27 x^{3}+y^{3}+z^{3}-9 x y z\\&=(3 x+y+z)\left(\begin{array}9 x^{2}+y^{2}+z^{2}-\\3 x y-y z-3 z x\end{array}\right)\end{align}\]

 Video Solution
Polynomials
Ex 2.5 | Question 11