NIMCET 2020 — Mathematics PYQ

Concept

- $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$
- If $a+b+c=0$, then $a^3+b^3+c^3=3abc$

Calculation

Since, $a+b+c=0$, we have $a^3+b^3+c^3=3abc$.

∴ The given expression is:

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

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

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