Question

prove that: (x+y)³+(y+z)³-3(x+y)(y+z)(z+x)=2(x³+y³+z³-3 xyz​

Answer 1

Answer:

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