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!

You can correct the error that Ryan points out. Instead, note that
$$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 8 - 6ab.$$
So,
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
becomes
$$a^5 + b^5 + 5ab(8 - 6ab) + 20a^2b^2 = 32.$$
Using $a^5 + b^5 = 32$, then expanding and collecting like terms,
$$40a^2b^2 - 10ab = 0 \implies ab = 0 \text{ or } ab = 4.$$
So, now we know that
$$(r - a)(r - b) = r^2 - (a + b)r + ab$$
equals either $r^2 - 2r + 4$, or $r^2 - 2r$, depending on the value of $ab$. Note that the former has no real roots (though it does have complex solutions, but that's whole other kettle of fish), but the latter does. The two numbers would have to be the roots of the former polynomial: $0$ and $2$. We must therefore have, putting the numbers in ascending order,
$$\sqrt[5]{16 - \sqrt{x}} = 0 \quad \sqrt[5]{16 + \sqrt{x}} = 2.$$
It's not hard to see that $\sqrt{x} = 16 \implies x = 256$.
Note: this is the same answer as your method. The reason is, because as it turned out $b = 0$, we actually had $(a + b)^3 = a^3 + b^3$ for that particular $a$ and $b$.
