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!
 A: Hint:
$$(x+1)^5-13(x+1)^3+36(x+1)=0$$
$$\left[(x+1)^4-13(x+1)^2+36\right](x+1)=0$$
$$\left((x+1)^2-9\right)\left((x+1)^2-4\right)(x+1)=0$$
A: Let $$(x+1)=t$$ Then $t$ satisfies the quintic $$t^5+36t=13t^3 \iff t(t^2-4)(t^2-9)=0 \iff t(t-2)(t+2)(t-3)(t+3)=0$$
Actually, in what you did your last step was invalid as you did not consider the case when you have that $x+1=0$.
A: 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.
A: 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
A: Assuming that you don't see the direct substitution $t=x+1$ for whatever reason, note that $(x+1)^5 +36x+36 - 13(x+1)^3$ is a monic polynomial with the free term $1+36-13=24\,$. By the rational root theorem, try the divisors of $24$ for possible rational roots, and enjoy your luck.
