<s>[INST] 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!
 [/INST] 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.
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