How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1)   = 13 (x +1)^3\\
(x+1)^4 +36         = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!

It's a simple problem. Well
$$(x+1)^5+36x+36=13(x+1)^3$$
$$\implies (x+1)^5–13(x+1)^3+36(x+1) = 0$$
Let $y=x+1$
$$\implies y^5-13y^3+36y = 0$$
$$\implies y(y^4-13y^2+36)= 0$$
$$\implies y(y^2-9)(y^2-4)= 0$$
$$\implies y = 0,\pm 2,\pm 3$$
$$\implies x+1 = 0,\pm 2,\pm 3$$
$$\implies x = -4, -3, -1, 1, 2$$
and we’re done.
