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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

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