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: $$ b = 0 $$ $$ \sqrt[5]{16-\sqrt x} = 0 $$ $$ 16 - \sqrt x = 0 $$ $$ x = 16^2 = 256 $$

When you went from : $$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$ to $$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$ You incorrectly assumed that $$a^3 + b^3 = (a+b)^3 = \biggl(\frac{32}{(a+b)^2}\biggr)$$ Therefore you cannot say that $$a^3 + b^3 = \frac{32}{2^2}$$