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

Question

If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$

Answer 1

$a+b+c=0\iff b+c=-a\implies (b+c)^2=a^2$

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

Actually, $b+c=-a\implies(b+c)^3=(-a)^3$

$\implies -a^3=b^3+c^3+3bc(b+c)=b^3+c^3+3bc(-a)\iff\sum a^3=3abc$

Answer 2

since $$LHS=\dfrac{a^2}{3bc}+\dfrac{b^2}{3ca}+\dfrac{c^2}{3ab}=\dfrac{a^3+b^3+c^3}{3abc}=\dfrac{3abc}{3abc}=1$$

I think this condition $a^3+b^3+c^3=3abc$ is redundant.because $$a+b+c=0\Longrightarrow a^3+b^3+c^3=3abc$$ because $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$

Answer 3

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