Question

28. Verify that
x3 + y2 + z3 - 3xyz
1
= 3 (x +3
(x + y + z) [(x - y)2 + (y – z)2 + (z – x)2]​

Answer 1

Answer:

I hope you understand the answer

Step-by-step explanation:

Answer

We have, L.H.S.=x

3+y^3+z^3 −3xyz

=(x+y+z)(x^2+y^2+z^2−xy−yz−zx)

[by poynomial identity]

=1/2(x+y+z)(2x2+2y^2+2z^2−2xy−2yz−2zx)

=1/2 [(x+y+z)(x−y)^2+(y−z)^2+(z−x)^2 ]

Hence proved.