Question:

Verify that x 3 + y 3 + z 3 – 3xyz = \(\frac{1}{2}\) (x + y + z) [(x - y)2 + (y - z)2 + (z - x)2].

Updated On: Nov 18, 2023
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Solution and Explanation

It is known that, 

x3 + y3 + z3 - 3xyz 

= (x + y + z) (x2 + y2 + z2 - xy - yz - zx) 

\(\frac{1}{2}\) (x + y + z) [2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx] 

=\(\frac{1}{2}\) (x + y + z) [(x2 + y2 - 2xy) + (y2 + z2 - 2yz) + (x2 + z2 -2zx)] 

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

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