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!

Spoiler alert: this is the solution.
Let's take $ a = \sqrt[5]{16+\sqrt x} $ and $ b = \sqrt[5]{16-\sqrt x} $ . Note that because $ \sqrt x \ge 0 $, $ a $ is always positive (technically greater than or equal to $ \sqrt[5]{16} $), but $ b $ can be positive, zero, or negative. We have:
$$ a + b = 2 $$
$$ (a + b)^5 = 32 $$
We also have:
$$ a^5 = 16 + \sqrt x $$
$$ b^5 = 16 - \sqrt x $$
$$ a^5 + b^5 = 32 $$
Aha! Let's see where this takes us:
$$ (a + b)^5 = a^5 + b^5 $$
$$ a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 = a^5 + b^5 $$
$$ 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 = 0 $$
For this sum to be zero, either all of its terms should be zero, or some of them should be positive and some others negative. The former case happens if $ b $ is zero (we noted earlier that $ a $ is always positive, so it can't be zero). The latter case can only happen if $ b $ is negative, because again as we noted earlier, $ a $ is always positive and cannot make negative terms. Let's keep that in mind and continue.
$$ 5ab(a^3 + 2a^2b + 2ab^2 + b^3) = 0 $$
The expression in the parentheses looks a bit like $ (a+b)^3 $, but not exactly! Let's add $ a^2b + ab^2 - a^2b - ab^2 $ to it:
$$ 5ab[(a+b)^3 - a^2b - ab^2] = 0 $$
$$ 5ab[(a+b)^3 - ab(a+b)] = 0 $$
$$ 5ab(a+b)[(a+b)^2-ab] = 0 $$
And recalling that $ a+b = 2 $:
$$ 10ab[4-ab]=0 $$
Now, for this equality to hold, we must have either $ ab = 4 $ or $ ab = 0 $.
But $ ab = 4 $ is not possible, because if it were so, considering that $ a $ is positive, $ b $ should necessarily be positive as well (otherwise $ ab $ would not be positive). And if both $ a $ and $ b $ were positive, then the sum $ 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 $ would necessarily be positive, whereas it should be zero.
So we are left with the conclusion that $ ab = 0 $. As we mentioned before, $ a $ is positive and not zero. So:
$$ b = 0 $$
$$ \sqrt[5]{16-\sqrt x} = 0 $$
$$ 16 - \sqrt x = 0 $$
$$ x = 16^2 = 256 $$
