An Algebra Problem by Percy Jackson

$\text{Identity used} \implies a^{3} + b^{3} + c^{3} = (a + b + c)(a^{2} + b^{2} + c^{2} -ab - bc - ca) + 3abc$

$\begin{aligned} \dfrac{a^{2}}{bc} + \dfrac{b^{2}}{ac} + \dfrac{c^{2}}{ab} &= \dfrac{a^{3}}{abc} + \dfrac{b^{3}}{abc} + \dfrac{c^{3}}{abc} \\ &= \dfrac{a^{3} + b^{3} + c^{3}}{abc} \\ &= \dfrac{(a + b + c)(a^{2} + b^{2} + c^{2} -ab - bc - ca) + 3abc}{abc} \\ &= \dfrac{0 \times (a^{2} + b^{2} + c^{2} -ab - bc - ca) + 3abc}{abc} \\ &= \dfrac{3abc}{abc} \\ &= \boxed{3} \end{aligned}$

In case you didn't know the identity...let $x$ be the required quantity. Multiplying by $abc$ , $abcx=a^3+b^3+c^3$

Substitute in $c=-(a+b)$ : $\begin{aligned} -ab(a+b)x &= a^3+b^3-(a+b)^3 \\ &=-3a^2 b - 3ab^2 \\ &=-3ab(a+b) \end{aligned}$

Dividing through by $-ab(a+b)$ gives $x=\boxed3$ . (There's no problem dividing through because we're told $a,b,c$ are non-zero.)

Using the identity:

$\begin{aligned} a^3+b^3+c^3 & = \blue{\cancel{(a+b+c)}^0}(a^2+b^2+c^2 - ab -bc-ca) + 3abc & \small \blue{\text{Since }a+b+c = 0} \\ a^3+b^3+c^3 & = 3abc \\ \frac {a^3+b^3+c^3}{abc} & = 3 \\ \implies \frac {a^2}{bc} + \frac {b^2}{ca} + \frac {c^2}{ab} & = \boxed 3 \end{aligned}$

LOL, I admit something, I substituted values for them to make my work easy :)

$a = 1; b = 2; c = -3$

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

$=\Large\frac{1^2}{2×-3}+\frac{2^2}{-1×3}+\frac{3^2}{1×2}$ (substitution)

$=\Large\frac{1}{-6}+\frac{4}{-3}+\frac{9}{2}$

$\boxed{=3}$