यदि $a$ और $b$ भिन्न-भिन्न पूर्णांक हों, तो सिद्ध कीजिए कि $\left(a^{n}-b^{n}\right)$ का एक गुणनखंड $(a-b)$ है, जबकि $n$ एक धन पूर्णांक है।

In order to prove that $(a-b)$ is a factor of $\left(a^{n}-b^{n}\right)$, it has to be proved that $a^{n}-b^{n}=k(a-b),$ where $k$ is some natural number

It can be written that, $a=a-b+b$

$\therefore a^{n}=(a-b+b)^{n}=[(a-b)+b]^{n}$

$ = {\,^n}{C_0}{(a - b)^n} + {\,^n}{C_1}{(a - b)^{n - 1}}b +  \ldots  + {\,^n}{C_{n - 1}}(a - b){b^{n - 1}} + {\,^n}{C_n}{b^n}$

$ = {(a - b)^n} + {\,^n}{C_1}{(a - b)^{n - 1}}b +  \ldots  + {\,^n}{C_{n - 1}}(a - b){b^{n - 1}} + {b^n}$

$\Rightarrow a^{n}-b^{n}=(a-b)\left[(a-b)^{n-1}+^{n} C_{1}(a-b)^{n-2} b+\ldots+^{n} C_{n-1} b^{n-1}\right]$

$\Rightarrow a^{n}-b^{n}=k(a-b)$

Where, $k = \left[ {{{(a - b)}^{n - 1}} + {\,^n}{C_1}{{(a - b)}^{n - 2}}b +  \ldots  + {\,^n}{C_{n - 1}}{b^{n - 1}}} \right]$ is a natural mumber

This shows that $(a-b)$ is a factor of $\left(a^{n}-b^{n}\right)$, where $n$ is a positive integer.