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If $n$ is even then show that $x-a $ is the factor of $x^n - a^n $.

My Attempt,

Let $P(x)= x^n-a^n $

$g (x) =x-a$

Comparing $g (x)$ with $x-A $ we get, $A=a$

By Remainder Theorem, $R=P(A)$

$R=P(a)$

$R=a^n - a^n$.

This is clearly equal to zero. Then what is the use of the condition given in the question that $n $ is even.

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    Lookup the [difference of two $n^{th}$ powers](https://en.wikipedia.org/wiki/Factorization#Sum.2Fdifference_of_two_nth_powers) identity (doesn't really matter whether $n$ is even or odd). – 2017-01-28
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    As for example, if $n$ is odd, (let $n=3$) then $R=a^3-a^3=0$.. – 2017-01-28
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    `Then what is the use of the condition` If the statement is true for all integer $n$ then it will be true for *even* $n$ in particular, but the condition is not necessary and, in fact, somewhat odd. On the other hand, if you had to prove that $x+a \mid x^n-a^n$, instead, then $n$ even would be required. – 2017-01-28
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    Whether $n $ is even or odd, it is true for pretty much the reason you gave. However if n is even then $x+a $ is a factor of $x^n-a^n $, which might not be true for n odd. So... weird question. Are you sure it wasn't meant to be x+a is a factor? – 2017-01-28
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    @fleablood, yeah I am sure – 2017-01-28
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    Well, the you show this the same way as showing that if $n$ is a perfect square (or a prime, or a composite, or a perfect number, or odd, or a multiple of 7) the result holds. "If n is even then.... what is the use of the condition given in the question that n is odd" Um, So which is it? n is even, or n is odd? You state both. – 2017-01-28
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    @fleablood, I am sorry for the typo. Its actually even. – 2017-01-28

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You could say $x^n - a^n = (x-a)P(x) + r$ where $P(x)$ is a polynomial of degree $n-1$ and $r$ is a real number. Letting $x=a$, you get $a^n - a^n = (a-a)P(x) + r$. Which simplifies to $r=0$. So, now, $x^n - a^n = (x-a)P(x)$. Hence $x-a$ is a factor of $x^n - a^n$.