<s>[INST] Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$
$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$
Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$
Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$
However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.
So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!
 [/INST] When you know $a+b$, you can express $a^n+b^n$ in terms of powers of $c=ab$. A few cases

*

*$a^2+b^2=(a+b)^2-2ab$

*$a^3+b^3=(a+b)^3-3ab(a+b)$
What about $a^5+b^5$? The relations are obtained from the expansion of $(a+b)^5$:
\begin{align}
(a+b)^5&=a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)\\
&=a^5+b^5+5ab((a+b)^3-3ab(a+b))+10a^2b^2(a+b)\\
&=a^5+b^5+5ab(a+b)^3-5a^2b^2(a+b)
\end{align}
In your case you get
$$
a^5+b^5=32-40c+10c^2
$$
which yields
$$
32=32-40c+10c^2
$$
hence $c=0$ or $c=4$.
The first case yields $16-\sqrt{x}=0$; the second case yields
$$
256-x=4^5
$$

Alternative approach. Since $a=\sqrt[5]{16+\sqrt{x}}\ne0$, we can set $t=b/a$ and from $(a+b)^5=a^5+b^5$, we get
$$
f(t)=(t+1)^5-t^5-1=0
$$
We have $f'(t)=5(t+1)^4-5t^4$, which vanishes only at $t=-1/2$, and $f(-1/2)<0$. Thus $f$ only vanishes twice, but one root is negative. Since $0$ is a root, it is the only nonnegative one. Thus $b=0$.
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