Verify that $ x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] $

To do:

We have to verify that \( x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \).

Solution:

We know that,

$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$

Therefore,

LHS $=x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)(x^2 + y^2 + z^2-xy-yz-zx)$

$= \frac{1}{2}(x+ y+ z)[2(x^2 + y^2 + z^2 - xy- yz - zx)]$

$= \frac{1}{2}(x+ y+ z)(2x^2 + 2y^2 + 2z^2 - 2xy- 2yz - 2zx)$

$= \frac{1}{2}(x+y+z)(x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx)$

$= \frac{1}{2}(x+y+ z)(x^2 + y^2 - 2xy+ y^2 + z^2 - 2yz+ x^2 + z^2 -2zx)$

$= \frac{1}{2}(x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2]$

$=$ RHS

Hence, $x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2]$.

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Updated on: 10-Oct-2022

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