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In the problem below I am unsure about the validity of the manipulation in Step 3 of the solution.

Tldr : How is $(1/x^2+1+x^2)^n = 1/x^{2n}(1 + x^2 + x^4)^n$? (I plugged in small numbers and it does show that the equation is valid but I do not understand how it was deduced)

The problem :

Given $(1+x+x^2)^n$ = $a_0 + a_1x + a_2x^2 + .... a_{2n}x^{2n}$ find $a_0^2 - a_1^2 + a_2^2 + ... a_{2n}^2$

The solution is as follows :

Step 1 : Let $LHS = (1+x+x^2)^n * (1- 1/x + 1/x^2)^n$

$i.e : = (a_0 + a_1x + a_2x^2 + .... a_{2n}x^{2n})(a_0 - a_1/x + a_2/x^2 +...+a_{2n}/x^{2n})$

This gives us the sum that has to be found on $LHS$ as the term independent of $x$.

Step 2 : Let $RHS = (1+x+x^2)^n * (1- 1/x + 1/x^2)^n$ = $[(1+x+x^2)(1- 1/x + 1/x^2)]^n$ = $(1/x^2 + 1 + x^2)^n$

This is where I don't understand how $1/x^{2n}$ can be factored out :

Step 3 : $RHS = 1/x^{2n}(1 + x^2 + x^4)^n$ = $1/x^{2n}(a_0 + a_1x^2 + a_2x^4 + ... + a_{2n}^{4n})$

This indicates that the term independent of $x$ on $RHS$ is $a_n$ which is thus equal to the term independent of $x$ on $LHS$, thus giving the solution.

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1 Answer 1

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$$\left(\frac1{x^2}+1+x^2\right)^n=\left(\frac1{x^2}(1+x^2+x^4)\right)^n =\left(\frac1{x^2}\right)^n(1+x^2+x^4)^n=\frac1{x^{2n}}(1+x^2+x^4)^n.$$

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