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Solution
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x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
First take L.H.S
(x+y+z)(x2+y2+z2−xy−yz−zx)
To multiply two polynomials, we multiply each monomial of one polynomial (with its sign) by each monomial (with its sign) of the other polynomial.
x.x2+x.y2+x.z2−x2y−xyz−x2z+y.x2+y.y2+y.z2−xy2−y2z−xyz+z.x2+z.y2+z.z2−xyz−yz2−xz2
= x3+xy2+xz2−x2y−x2y+yx2+y3−xy2−y2z+x2z+y2z+z3−yz2−xz2−3xyz
= x3+y3+z3−3xyz
L.H.S = R.H.S
x3+y3+z3−3xyz=x3+y3+z3−3xyz
Hence x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx) is proved.
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