Ex.2.5 Q12 Polynomials Solution - NCERT Maths Class 9

Go back to  'Ex.2.5'

Question

Verify that:

\[\begin{align}&x^{3}+y^{3}+z^{3}-3 x y\\&=\frac{1}{2}(x+y+z)\left[\begin{array}((x-y)^{2}+(y-z)^{2}\\+(z-x)^{2}\end{array}\right]\end{align}\]

 Video Solution
Polynomials
Ex 2.5 | Question 12

Text Solution

Reasoning:

Identity:

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

Steps:

Taking RHS

\[\begin{align}&=\frac{1}{2}(x+y+z)\left[\begin{array}((x-y)^{2}+(y-z)^{2}\\+(z-x)^{2}\end{array}\right] \\ &=\frac{1}{2}(x+y+z)\left[\begin{array}((x^{2}-2 x y+y^{2})+\\(y^{2}-2 y z+z^{2})+\\(z^{2}-2 z x+x^{2})\end{array}\right]\\ &=\frac{1}{2}(x+y+z)\left[\begin{array}(2 x^{2}+2 y^{2}+2 z^{2}-\\2 x y-2 y z-2 z x\end{array}\right] \\ &=\frac{1}{2}(x+y+z)(2)\left[\begin{array}(x^{2}+y^{2}+z^{2}-\\x y-y z-z x\end{array}\right]\\&=\!\begin{Bmatrix}\!x\left[x^{2}+y^{2}+z^{2}-x y-y z-z x\right]\!+\!\!\!\\\!y\left[x^{2}+y^{2}+z^{2}-x y-y z-z x\right]\!+\!\!\!\\z\left[x^{2}+y^{2}+z^{2}-x y-y z-z x\right]\end{Bmatrix}\\ &=\left[\begin{array}(x^{3}+x y^{2}+x z^{2}-x^{2} y-x y z-\\x^{2} z+x^{2} y +y^{3}+y z^{2}-x y^{2}-\\y^{2} z-x y z+z x^{2}+y^{2} z+z^{3}-\\x y z-y z^{2}-x z^{2} \end{array}\right] \\ &=x^{3}+y^{3}+z^{3}-3 x y z=\mathrm{LHS}\end{align}\]