2021/024) if $a^3+b^3+c^3=3abc$ and a,b, c are different then prove that $\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac}+\frac{(a+b)2}{3ab}=1$

We have $a^3+b^3+c^3=3abc$ so a=b=c or a+b+c = 0

As a,b,c are different so a+b + c = 0

Hence

$\frac{(b+c)^2}{3bc}= \frac{(-a)^2}{3bc}=\frac{a^3}{3abc}$

Similarly we have 

$\frac{(c+a)^2}{3ac}= \frac{b^3}{3abc}$

$\frac{(a+b)^2}{3ab}= \frac{c^3}{3abc}$ 

Adding these 3 we get

$\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}= \frac{a^3+b^3+c^3}{3abc} = \frac{3abc}{3abc}=1$