$$5x^3+ax^2+8 = 0 = 5x^3 + 8x^2 + a $$ Subtracting $5x^3$ we get $$8x^2+a=ax^2+8$$ and rearranging terms we get $$(8-a)(x^2-1)=0$$ So $a=8$ or $x=\pm 1$ Case 1: x = 1 $$5(1)^3 + a(1)^2 + 8 = 0$$ $$a+13 = 0$$ $$a = -13$$ $$\boxed{(-13,1)}$$ Case 2: x = -1 $$5(-1)^3 + a(-1)^2 + 8 = 0$$ $$a+3 = 0$$ $$a = -3$$ $$\boxed{(-3,-1)}$$ Case 3: a=8 $$5x^3+8x^2+8 = 0$$ $$(x+2)(5x^2-2x+4)$$ So $x = -2$ or $5x^2-2x+4=0$, which has no real solutions. $$\boxed{(8,-2)}$$ All Three: $\boxed{(-13,1),(-3,-1),(8,-2)}$