Algebraic Identities — Maths STD 9 — Question

$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
If $\mathrm{a}+\mathrm{b}+\mathrm{c}=0$, then
$a^3+b^3+c^3-3 a b c=0$
$\Rightarrow a^3+b^3+c^3=3 a b c$
Now, consider $\frac{\mathrm{a}^2}{\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{ca}}+\frac{\mathrm{c}^2}{\mathrm{ab}}$
Multiplying dividing by a. b. and c in $\frac{\mathrm{a}^2}{\mathrm{bc}} \cdot \frac{\mathrm{b}^2}{\mathrm{ca}}$ and $\frac{\mathrm{c}^2}{\mathrm{ab}}$ respectively. we get
$\frac{\mathrm{a}^3}{\mathrm{abc}}+\frac{\mathrm{b}^3}{\mathrm{bca}}+\frac{\mathrm{c}^3}{\mathrm{cab}}$
$=\frac{\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3}{\mathrm{abc}}$
$=\frac{3 \mathrm{abc}}{\mathrm{abc}} \ldots[\text { From (1)] }$
$=3$
Hence, correct option is $(d)$.