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!

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
$$a^3 + b^3 = \frac{32}{2^2}$$
