If ${\left(1+x+x^2\right)}^n=a_0+a_{1}x+a_2{x}^2+\dots +a_{2n}{x}^{2n}$  then $a_0^2-a_1^2+a_2^2-a_3^2+\dots +a_{2n}^2=$ 

We have,

${\left(1+x+x^2\right)}^n=a_0+a_{1}x+a_2{x}^2+\dots +a_{2n}{x}^{2n}$

Replacing $x\text{ by }-\frac{1}{x}\text{, we get }$

${\left(1-\frac{1}{x}+\frac{1}{x^2}\right)}^n=a_0-\frac{a_1}{x}+\frac{a_2}{x_2}-\frac{a_3}{x^3}+\dots +\frac{a_{2n}}{x^{2n}}$

Multiplying both sides by $x^{2n}$ we get

$\begin{array}{r}{\left(x^2-x+1\right)}^n=a_0{x}^{2n}-a_1{x}^{2n-1}+a_2{x}^{2n-2}-a_3{x}^{2n-3}+\dots +a_2\\ \Rightarrow {\left(1-x+x^2\right)}^n=a_0{x}^{2n}-a_1{x}^{2n-1}+a_2{x}^{2n-2}-a_3{x}^{2n-3}+\end{array}$

Multiplying (i) and (ii), we get

$\begin{array}{l}{\left(1+x+x^2\right)}^n{\left(1-x+x^2\right)}^n\\ =\left(a_0+a_{1}x+a_2{x}^2+\dots +a_{2n}{x}^{2n}\right)\\ \times \left(a_0{x}^{2n}-a_1{x}^{2n-1}+a_2{x}^{2n-2}+\dots +a_2\right)\\ {\left\{{\left(1+x^2\right)}^2-x^2\right\}}^n=\left\{a_0+a_{1}x+a_2{x}^2+\dots +a_{2n}{x}^{2n}\right\}\\ \times \left\{a_0{x}^{2n}-a_1{x}^{2n-1}+a_2{x}^{2n-2}+\dots +a_{2n}\right.\\ {\left(1+x^2+x^4\right)}^n=\left(a_0+a_{1}x+a_2{x}^2+\dots +a_{2n}{x}^{2n}\right)\\ \times \left(a_0{x}^{2n}-a_1{x}^{2n-1}+a_2{x}^{2n-2}+\dots +a_{2n}\right)\dots \left(\text{ iii }\right)\end{array}$

Replacing x by $x^2$ in (i), we get

${\left(1+x^2+x^4\right)}^n=\sum _{r=0}^{2n} a_r{x}^{2r}....\left(iv\right)$

$\sum _{r=0}^{2n} a_r{x}^{2r}=\left(a_0+a_1{x}^2+a_2{x}^4+\dots +a_{2n}{x}^{4n}\right)$

compare coefficient of $x^{2n} in \left(iii\right),\left(iv\right)$