<s>[INST] Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\\-(t-2)(t+2)^3+(t+2)^4\right]=242t$$ Let $A=(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-(t-2)(t+2)^3+(t+2)^4.$
Then $$A=(t-2)^4-(t-2)^2(t^2-4+t^2+4t+4)-(t+2)^3(2-t+t+2)\\=(t-2)^4-2t(t+2)(t-2)^2-4(t+2)^3.$$
 [/INST] We have
$$0=(x-1)^5+(x+3)^5-242(x+1)=2(x^2 + 2x + 42)(x + 2)(x + 1)x
$$
by applying the rational root theorem to the polynomial equation
$$
2x^5 + 10x^4 + 100x^3 + 260x^2 + 168x=0,
$$
which yields the linear factors $x$, $x+1$ and $x+2$.
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