If $a$ and $b$ are distinct integers,prove that $a-b$ is a factor of $a^{n}-b^{n}$,whenever $n$ is a positive integer.

(N/A) To prove that $(a-b)$ is a factor of $(a^{n}-b^{n})$,we must show that $a^{n}-b^{n} = k(a-b)$,where $k$ is an integer.
We can write $a$ as $(a-b)+b$.
Therefore,$a^{n} = ((a-b)+b)^{n}$.
Using the Binomial Theorem,$a^{n} = \sum_{r=0}^{n} {^{n}C_{r}} (a-b)^{n-r} b^{r}$.
Expanding this,$a^{n} = {^{n}C_{0}}(a-b)^{n} + {^{n}C_{1}}(a-b)^{n-1}b + \dots + {^{n}C_{n-1}}(a-b)b^{n-1} + {^{n}C_{n}}b^{n}$.
Since ${^{n}C_{0}} = 1$ and ${^{n}C_{n}} = 1$,we have $a^{n} = (a-b)^{n} + {^{n}C_{1}}(a-b)^{n-1}b + \dots + {^{n}C_{n-1}}(a-b)b^{n-1} + b^{n}$.
Subtracting $b^{n}$ from both sides,$a^{n}-b^{n} = (a-b)^{n} + {^{n}C_{1}}(a-b)^{n-1}b + \dots + {^{n}C_{n-1}}(a-b)b^{n-1}$.
Factoring out $(a-b)$,we get $a^{n}-b^{n} = (a-b) [ (a-b)^{n-1} + {^{n}C_{1}}(a-b)^{n-2}b + \dots + {^{n}C_{n-1}}b^{n-1} ]$.
Let $k = [ (a-b)^{n-1} + {^{n}C_{1}}(a-b)^{n-2}b + \dots + {^{n}C_{n-1}}b^{n-1} ]$,which is an integer.
Thus,$a^{n}-b^{n} = k(a-b)$,proving that $(a-b)$ is a factor of $a^{n}-b^{n}$ for any positive integer $n$.