<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] You could also solve it in a much simpler way, like this:
$$
\begin{align*}
(x - 1)^{5} + (x + 3)^{5} &= 242 \cdot (x + 1) \quad\mid\quad -(242 \cdot (x + 1))\\
(x - 1)^{5} + (x + 3)^{5} - 242 \cdot (x + 1) &= 0\\
2 \cdot x^{5} + 10 \cdot x^{4} + 100 \cdot x^{3} + 260 \cdot x^{2} + 168 \cdot x &= 0 \quad\mid\quad \div 2\\
x^{5} + 5 \cdot x^{4} + 50 \cdot x^{3} + 130 \cdot x^{2} + 84 \cdot x &= 0\\
x \cdot (x^{4} + 5 \cdot x^{3} + 50 \cdot x^{2} + 130 \cdot x + 84) &= 0 \quad\mid\quad \text{factor into four terms}\\
x \cdot (x + 1) \cdot (x + 2) \cdot (x^{2} + 2 \cdot x + 42) &= 0 \quad\mid\quad \text{use the zero product theorem}\\
\\
x &= 0 \tag{1.}\\
x + 1 &= 0 \quad\mid\quad -1 \tag{2.}\\
x + 2 &= 0 \quad\mid\quad -2 \tag{3.}\\
x^{2} + 2 \cdot x + 42 &= 0 \quad\mid\quad -41 \tag{4.}\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
x^{2} + 2 \cdot x + 1 &= -41 \tag{4.}\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
(x + 1)^{2} &= -41 \quad\mid\quad \sqrt{~~~} \tag{4.}\\
x + 1 &= \pm\sqrt{41} \cdot \mathrm{i} \quad\mid\quad -1\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
x &= -1 \pm \sqrt{41} \cdot \mathrm{i} \tag{4.}\\
\end{align*}
$$
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