Solve the equation $$(x+1)^4=2(1+x^4)$$ The most intuitive approach for me was to use the formula $$(a+1)^4=a^4+4a^3+6a^2+4a+1,$$ so our equation is $$x^4+4x^3+6x^2+4x+1-2-2x^4=0\\-x^4+4x^3+6x^2+4x-1=0$$ $\pm1$ aren't solutions, so this equation does not have whole roots.
Another thing I tried is to factor $$1+x^4\ne(x+1)(x^3-x^2+x-1)=x^4-1,$$ but then I remembered it is for odd $n$. I don't know what else to try.
Based on the coefficients and the polynomial $ x^4 - 4 \, x^3 - 6 \, x^2 - 4 \, x + 1 = 0$ try a product of the type $$( x^2 - 2 \, a \, x + 1)(x^2 - 2 \, b \, x + 1) = 0$$ which gives $$ a + b = 2 \hspace{5mm} \text{and} \hspace{5mm} a \, b = -2.$$ The roots of the original polynomial take the form $$ x \in \left\{ a + \sqrt{a^2 - 1}, \, a - \sqrt{a^2 - 1}, \, b + \sqrt{b^2 - 1}, \, b - \sqrt{b^2 - 1}\;\right\}.$$ What is left is to solve for $a$ and $b$.
