Question

If $x=b+c-a,y=c+a-b,z=a+b-c$, shew that $x^3+y^3+z^3-3xyz=4(a^3+b^3+c^3-3abc)$.

Answer 1

Given that:

$x=b+c-a,y=c+a-b,z=a+b-c$

To Show:

$x^3+y^3+z^3-3xyz=4(a^3+b^3+c^3-3abc)$

Proof:
$x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\left\{\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\}$

Here,

$x+y+z=b+c-a+c+a-b+a+b-c=a+b+c,$

$x-y=b+c-a-c-a+b=2(b-a),$

Similarly,

$y-z=2(c-b)$ and $z-x=2(a-c)$

Now,

L.H.S$=x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\left\{\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\}$

$=\frac{1}{2}(a+b+c)\left\{4\right(b-a{)}^2+4(c-b{)}^2+4(a-c{)}^2\}$

$=4\times \frac{1}{2}(a+b+c)\left\{\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}$

$=4(a^3+b^3+c^3-3abc)$

$=$R.H.S

Hence, proved.