A Triangle Problem!

$\dfrac {a^3+b^3+c^3}{a+b+c} = c^2 \\ a^3+b^3+c^3 = (a+b+c)c^2 \\ c^2 = \dfrac {a^3+b^3}{a+b} = a^2 - ab + b^2 \\ c^2 = a^2 + b^2 - 2ab\cos C \\ \therefore \cos C = \dfrac 12 \implies C = \dfrac {\pi}3 \\ S_{\triangle} = \dfrac 12 ab\sin C \\ \dfrac {S_{\triangle}}{ab} = \dfrac 12 \sin C = \dfrac 12 \sin {\dfrac {\pi}3} = \dfrac {\sqrt 3}4 \\ \implies \omega = 3, λ = 4, \omega + λ = \boxed {7}$

$\cfrac{a^3 + b^3 + c^3}{a + b + c} = c^2$ rearranges to $c^2 = a^2 - ab + b^2$ .

By a variation of Heron's formula , $A_{\triangle ABC} = \cfrac{1}{4}\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2}$ , which after substituting $c^2 = a^2 - ab + b^2$ simplifies to $A_{\triangle ABC} = \cfrac{\sqrt{3}}{4}ab$ .

Therefore, $\cfrac{A_{\triangle ABC}}{ab} = \cfrac{\frac{\sqrt{3}}{4}ab}{ab} = \cfrac{\sqrt{3}}{4}$ , so $\omega = 3$ , $\lambda = 4$ , and $\omega + \lambda = 3 + 4 = \boxed{7}$ .

Let $\dfrac{a^3 + b^3 + c^3}{a + b + c} = c^2$

Using the law of cosines $\implies c^2 = a^2 + b^2 - 2ab\cos(\alpha) \implies$

$a^3 + b^3 + c^3 = c^2(a + b + c) = (a + b + c)(a^2 + b^2 - 2ab\cos(\alpha)) =$

$a^3 + b^3 + ab^2 + ba^2 + ca^2 + cb^2 - 2(a^2b + ab^2 + abc)\cos(\alpha)) \implies$

$c^3 - c(a^2 + b^2) = ab^2 + ba^2 - 2(a^2b + ab^2 + abc)\cos(\alpha)) \implies$

$c(c^2 - (a^2 + b^2)) = ab^2 + ba^2 - 2(a^2b + ab^2 + abc)\cos(\alpha) \implies$

$c(-2ab\cos(\alpha)) = -2abc\cos(\alpha) = ab^2 + ba^2 - 2(a^2b + ab^2 + abc)\cos(\alpha) \implies$

$ab^2 + ba^2 - 2(a^2b + ab^2)\cos(\alpha) = 0 \implies (ab^2 + ba^2)(1 - 2\cos(\alpha)) = 0 \implies$

$ab(a + b)(1 - 2\cos(\alpha)) = 0$ and $a > 0$ and $b > 0 \implies 1 - 2\cos(\alpha) = 0 \implies$

$\cos(\alpha) = \dfrac{1}{2} \implies \alpha = 60^{\circ} \implies$

$A_{\triangle{ABC}} = \dfrac{1}{2}ab\sin(60^{\circ}) = \dfrac{\sqrt{3}}{4}ab \implies \dfrac{A_{\triangle{ABC}}}{ab} = \dfrac{\sqrt{3}}{4} = \dfrac{\sqrt{\omega}}{\lambda}$

$\implies \omega + \lambda = \boxed{7}$ .