Question

Show that the product of $a^3+b^3+c^3-3abc$ and $x^3+y^3+z^3-3xyz$ can be put into the form $A^3+B^3+C^3-3ABC$.

Answer 1

Let the product of $a^3+b^3+c^3-3abc$ and $x^3+y^3+z^3-3xyz$ as $(a+b+c)(a+wb+w^{2}c)(a+w^{2}b+wc)\times (x+y+z)(x+wy+w^{2}z)(x+w^{2}y+wz)$.
By taking six factors in the pair;
$(a+b+c)(x+y+z);(a+wb+w^{2}c)(x+wy+w^{2}z);(x+w^{2}y+wz)(a+w^{2}b+wc)$
BY using partial products concept, we get
$(a+b+c)(a+wb+w^{2}c)(a+w^{2}b+wc)\times (x+y+z)(x+wy+w^{2}z)(x+w^{2}y+wz)$

$=(A+B+C)(A+wB+w^{2}C)(A+w^{2}B+WC)$
By comparing equal terms of $w,w^2$ on both sides, we get the values of $A,B$ and $C$ as
$A=ax+by+cz,B=bx+cy+az,C=cx+ay+bz$
Thus, we can write the product of $(A+B+C)(A+wB+w^{2}C)(A+w^{2}B+wC)=A^3+B^3+C^3-3ABC$