Question

verify that x3+y3+z3-3xyz=1/2(x+y+z)[(x-y)2+(z-x)2]​

Answer 1

$\huge\boxed{\purple{\bf{Question}}}$

Verify

$\sf x{}^{3}+y{}^{3}+z{}^{3}-3xyz=\frac{1}{2}(x+y+z)[(x-y){}^{2}+(y-z){}^{2}+(z-x){}^{2}]$

$\huge\mathbb{\underline{SOLUTION}}$

to prove that

$\sf {x}^{3}+{y}^{3}+z{}^{3}-3xyz=(x+y+z)(x{}^{2}+y{}^{2}+z{}^{2}-xy-yz-zx]$

$\bf{\underline{\underline{Now\: according\: to \:Question:-}}}$

$\sf L.H.S = RH.S$

$\large\mathbb{VERIFICATION}$

$\sf x{}^{3}+y{}^{3}+z{}^{3}-xyz=\frac{1}{2}(x+y+z)((x-y){}^{2}+(y-z){}^{2}+(z-y){}^{2})\\ \sf\implies R.H.S\\ \sf\leadsto \frac{1}{2}×(x+y+z)(x{}^{2}+y{}^{2}-2xy+(y{}^{2}+z{}^{2}-2yz+z{}^{2}+x{}^{2}-2zx)\\ \sf\leadsto \frac{1}{2}×(x+y+z)(2x{}^{2}+2y{}^{2}+2z{}^{2}-2xy-2yz-2zx)\\ \sf\leadsto \frac{1}{\cancel2}×(x+y+z) ×\cancel2(x{}^{2}-y{}^{2}-z{}^{2}-xy-yz-zx)\\ \sf\leadsto (x+y+z)(x{}^{2}+y{}^{2}+z{}^{2}-xy-yz-zx)$

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

L.H.S =R.H.S