Solutions for quartic

Question

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.

Answer 1

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 $$x_1=\frac{1}{2} \left(a+b-\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_2=\frac{1}{2} \left(a+b+\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_3=\frac{1}{2} \left(a+b-\sqrt{2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_4=\frac{1}{2} \left(a+b+\sqrt{2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$

Answer 2

let $u=x-\dfrac{a+b}{2},t=\dfrac{a-b}{2} \to x-a=u-t,x-b=u+t$

$(x-a)^4+(x-b)^4=c \iff (u-t)^4+(u+t)^4=c \iff ((u-t)^2)^2-2(u-t)^2(u+t)^2+((u+t)^2)^2+2(u-t)^2(u+t)^2=c \iff ((u-t)^2-(u+t)^2)^2+2(u^2-t^2)^2-c=0 \iff 16u^2t^2+2u^4+2t^4-4t^2u^2-c=0 \iff u^4+6u^2t^2+t^4-\dfrac{c}{2}=0$