Solve for $x$ in $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$ Solve for $x$ in $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
i have done in a lengthy way:
By inspection we observe that $x=41$ and $x=-24$ are the solutions
we have
$$\sqrt[4]{57-x}=5-\sqrt[4]{x+40}$$ squaring both sides we get
$$\sqrt{57-x}=25+\sqrt{x+40}-10 \sqrt[4]{x+40}$$ that is
$$\sqrt{57-x}-25=\sqrt{x+40}-10 \sqrt[4]{x+40}$$ again squaring both sides we get
$$682-x-50\sqrt{57-x}=x+40+100\sqrt{x+40}-20(x+40)^{\frac{3}{4}}$$
i got messed up here any better way and just a hint please

$$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
Let $y=\sqrt[4]{57-x}$, and $z=\sqrt[4]{x+40}$
Then
$$y\ge 0, z \ge 0, y+z=5$$
$$x=57- y^4=z^4-40$$
Thus
$$y^4+ (5-y)^4-97$$
$$=2(y-3)(y-2)(y^2-5y+44)=0$$
Thus $y=2$ or $y=3$, which means $x=41$ or $x=-24$
We are done.
