$$(x+1)(x-1)(x^2+1)(x^4+1) = 0$$
Answer
$$ \begin{matrix}x_1 = 1 & x_2 = -1 & x_3 = 0.70711+0.70711i \\[1 em] x_4 = 0.70711-0.70711i & x_5 = -0.70711+0.70711i & x_6 = -0.70711-0.70711i \\[1 em] x_7 = 1i & x_8 = -i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x+1)(x-1)(x^2+1)(x^4+1) &= 0&& \text{simplify left side} \\[1 em](x^2-x+x-1)(x^2+1)(x^4+1) &= 0&& \\[1 em](x^2-1)(x^2+1)(x^4+1) &= 0&& \\[1 em](x^4+x^2-x^2-1)(x^4+1) &= 0&& \\[1 em](x^4-1)(x^4+1) &= 0&& \\[1 em]x^8+x^4-x^4-1 &= 0&& \\[1 em]x^8+x^4-x^4-1 &= 0&& \\[1 em]x^8-1 &= 0&& \\[1 em] \end{aligned} $$
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