Question

If $a,b,c$ are all non-zero and $a+b+c=0$, prove that $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3$

Answer 1

Given,

$a+b+c=0$

or, $a+b=-c$

cubing both sides we get,

$(a+b{)}^3=-c^3$

$a^3+b^3+3ab(a+b)=-c^3$ [$\because (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$]

$a^3+b^3-3abc=-c^3$ [Since $a+b=-c$]

$a^3+b^3+c^3=3abc$.......(1).

Now,

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

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

$=\frac{3abc}{abc}$ [Using (1)]

$=3$.