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
 A: $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
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.
A: Firstly, I would recommend you to make the following substitution $a = \sqrt[4]{57 - x}$ and $b = \sqrt[4]{x+40}$. Thus we have $a + b = 5$ and $a^{4} + b^{4} = 57 + 40 = 97$. Therefore:
\begin{align*}
a^{4}+b^{4} & = (a^{2}+b^{2})^{2} - 2a^{2}b^{2} = [(a+b)^{2} - 2ab]^{2} - 2a^{2}b^{2} = (5^{2} - 2ab)^{2} - 2a^{2}b^{2}\\
& = (25 - 2ab)^{2} - 2a^{2}b^{2} = 625 - 100ab + 2a^{2}b^{2} = 97
\end{align*}
Henceforward we shall also agree that $\alpha = a + b$ and $\beta = ab$. According to the last relation, we obtain that $2\beta^{2} - 100\beta + 528 = 0$, whose solutions are $\beta_{1} = 6$ and $\beta_{2} = 44$. After solving the corresponding collection of systems of equations, we obtain the next solution set $S = \{-24,41\}$.
A: 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.
