Question

If $a^3+b^3+c^3=3abc$ and $a+b+c=0$, then find $\frac{{\left(b+c\right)}^2}{3bc}+\frac{{\left(c+a\right)}^2}{3ac}+\frac{{\left(a+b\right)}^2}{3ab}$

• A. $1$
• B. $-2$
• C. $2$
• D. $0$

Answer 1

The correct option is A $1$

Given :

$a^3+b^3+c^3=3abc$

$a+b+c=0$

$\therefore b+c=-a,c+a=-b,a+b=-c$

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

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

= $\frac{a^3+b^3+c^3}{3abc}$

= $\frac{3abc}{3abc}$

= $1$