Question

Prove the following identities:
$\sum (b-c{)}^3(b+c-2a)=0$; hence deduce $\sum (\beta -\gamma )(\beta +\gamma -2a{)}^3=0$.

Answer 1

$\Rightarrow$ Let $b-c=x,c-a=y,a-b=z$

We have to show that

$\sum x^3(y-z)=0$ when $x+y+z=0$

So, $x^3(y-z)+y^3(z-x)+z^3(x-y)$

can be reduced as;

$=k(y-z)(z-x)(x-y)(x+y+z)$

Hence due to the factor $x+y+z=0$, we have

$\sum x^3(y-z)=0$

or $\sum {(b-c)}^3(b+c-2a)=0$

Similarly for nect equation;-

Let $x=\beta +\gamma -2\alpha ;y=\gamma +\alpha -2\beta ;z=\alpha +\beta -2\gamma$

Hence to show;-

$\sum x^3\frac{(y-2)}{3}=0$ when $x+y+z=0$

So hence it can be proved as earlier