Solutions for quartic Suppose I have an equation in the form $(x-a)^4 + (x-b)^4 = c$. What is a clever way to find all four solutions? I have tried expanding and then used long division. However, I believe a better way is to reduce to cubic, then to quadratic. I don't know how to do this. 

The solutions of $$(x-a)^4 + (x-b)^4 = c$$ are less complex that we could think. They are, after some tedious simplifications given by $$x_1=\frac{1}{2} \left(a+b-\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_2=\frac{1}{2} \left(a+b+\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_3=\frac{1}{2} \left(a+b-\sqrt{2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$
$$x_4=\frac{1}{2} \left(a+b+\sqrt{2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$
