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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.
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Verified Answer
Now, $a=a+b-b=b+(a-b)$
$\begin{aligned}
a^n &=\{b+(a-b)\}^n \\
&=b^n+{ }^n C_1 b^{n-1}(a-b)+{ }^n C_2 b^{n-2}(a-b)^2
\end{aligned}$
$+\ldots \ldots \ldots \ldots+(a-b)^n$
or $a^n-b^n={ }^n C_1 b^{n-1}(a-b)+{ }^n C_2(a-b)^2 b^{n-2}$ $+\ldots \ldots \ldots \ldots+(a-b)^n$
$\begin{aligned}=(a-b)\left[{ }^n C_1 b^{n-1}+{ }^n C_2 b^{n-2}(a-b)\right.\\ &\left.+\ldots \ldots .+(a-b)^{n-1}\right] \end{aligned}$
Thus, $(a-b)$ is a factor of $\left(a^n-b^n\right)$
$\begin{aligned}
a^n &=\{b+(a-b)\}^n \\
&=b^n+{ }^n C_1 b^{n-1}(a-b)+{ }^n C_2 b^{n-2}(a-b)^2
\end{aligned}$
$+\ldots \ldots \ldots \ldots+(a-b)^n$
or $a^n-b^n={ }^n C_1 b^{n-1}(a-b)+{ }^n C_2(a-b)^2 b^{n-2}$ $+\ldots \ldots \ldots \ldots+(a-b)^n$
$\begin{aligned}=(a-b)\left[{ }^n C_1 b^{n-1}+{ }^n C_2 b^{n-2}(a-b)\right.\\ &\left.+\ldots \ldots .+(a-b)^{n-1}\right] \end{aligned}$
Thus, $(a-b)$ is a factor of $\left(a^n-b^n\right)$
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