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

Since you found two roots you may try to argue that there are no more roots.
One way to do this is study the function $f(x)=(57-x)^{1/4}+(x+40)^{1/4}$. 
Now, 
we can calculate $$f''(x)= \dfrac{3}{16} \left (-\dfrac{1}{(57 - x)^{7/4}} - \dfrac{1}{(40 + x)^{7/4}} \right )<0.$$
Therefore $f$ cannot have three roots in $(-40,57)$, since that would give two roots for $f'$ contradicting the fact that $f'$ is decreasing.
