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!

Continue from your last step.
Put $(x + 1)^2 = z$
We have,
$z^2 + 36 = 13z$
$z^2 - 13z + 36 = 0$
$z^2 - 9z - 4z  + 36 = 0$
$z(z - 9) - 4(z - 9) = 0$
$(z - 9)(z - 4) = 0$
When z = 9
$(x + 1)^2 = 9$
$(x + 1) = \pm 3$
x = 2 or -4
When z = 4 
$(x + 1)^2 = 4$
$(x + 1) = \pm 2$
x = 1 or -3
x = -4, -3, 1, 2
