Find the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2}- \dots+a_{2 n}^{2}$ Let n be a positive integer and $$\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n}$$
then the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2} - \dots+a_{2 n}^{2}$ is
My approach:-
Replacing $x$ by $(-1 / x),$ we get
$$
\begin{array}{r}
\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{n}=a_{0}-\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots-a_{2 n-1} \cdot \frac{1}{x^{2 n-1}}+\frac{a_{2 n}}{x^{2 n}} \\
\text { or, }\left(1-x+x^{2}\right)^{n}=a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+a_{2 n}..... \tag{1}
\end{array}
$$
And given $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n} \ldots \ldots \ldots \ldots \ldots . \tag{2}.$
Multiplying corresponding sides of (1) and $(2),$ we have
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\right) \times\left(a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+\right.
$$
$\left.a_{2 n}\right) \ldots \ldots...\tag{3}$
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x^{2}+a_{2} x^{4}+\cdots+a_{n} x^{2n}+\cdots+a_{2 n} x^{4 n}\right) \ldots \ldots\tag{4}
$$
Equating coefficient of $x^{2 n}$ on both sides of (3) and (4)
$$
a_{0}^{2}-a_{1}^{2}+a_{2}^{2} -\cdots +a_{2 n}^{2}=a_{n}
$$
But this method seems  very tedious to me.
Any other approach would be greatly appreciated

The following is essentially the idea in your proof which is conceptually simple. One has by the given, $$(1+x+x^2)^n=a_0+a_1x+\cdots+a_{2n}x^{2n}.\quad (1)$$
Replacing $x$ by $1/x$ and multiplying by $x^{2n}$ in (1), one sees that $$a_k=a_{2n-k}, 0\leq k\leq 2n.\quad (1)$$
Replacing $x$ by $-x$ in (1), one has $$(1-x+x^2)^n=a_0-a_1x+\cdots+a_{2n}x^{2n}.\quad (2)$$
Replacing $x$ by $x^2$ in (1), one has $$(1+x^2+x^4)^n=a_0+a_1x^2+\cdots a_n x^{2n}+\cdots+a_{2n}x^{4n}.\quad (3)$$
Since $1+x^2+x^4=(1+x+x^2)(1-x+x^2)$, multiplying (2) and (1) and comparing coefficients of $x^{2n}$ with (3), one has $$a_0a_{2n}-a_1a_{2n-1}+a_2a_{2n-2}+\cdots+a_{2n}a_0=a_n,$$ which after applying (1) yields $$a_0^2-a_1^2+a_2^2-\cdots+a_{2n}^2=a_n,$$ as required.
