Question

If $a+b+c=0$, and $x+y+z=0$, show that
$4{(ax+by+cz)}^3-3(ax+by+cz)(a^2+b^2+c^2)(x^2+y^2+z^2)-2(b-c)(c-a)(a-b)$

$(y-z)(z-x)(x-y)=54abcxyz$.

Answer 1

Suppose that $a=0$ in which case $c=-b$ then the left hand side becomes,

$4b^3{(y-z)}^3-6b^3(y-z)(x^2+y^2+z^2)-4b^3(y-z)(z-x)(x-y)$

$=2b^3(y-z)\{2{(y-z)}^2-3(x^2+y^2+z^2)+2(x-y)(x-z)\}$

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

$=-2b^3(y-z){(x+y+z)}^2$

$=0,\because x+y+z=0$

Hence the left hand side vanishes when $a=0;$ and similarly it vanishes when $b=0,c=0,x=0,y=0,z=0$ and thus may be put equal to $kabcxyz$

To find k, we put $a=1,b=1,c=-2,x=1,y=1,z=-2$ thus;

$4k=4\times 6^3-3\times 6\times 6\times 6$

$4k=6^3$

$k=\frac{6^3}{4}=54$