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# Prove that ${{x}^{n}}-{{y}^{n}}$ is divisible by $x-y$ .

Last updated date: 19th Jul 2024
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Hint: As per the question, let us assume that $P\left( n \right)={{x}^{n}}-{{y}^{n}}$ is divisible by $x-y$ and first we will prove that P (1) is divisible by $x-y$. And now if it is true for P(k+1) then it will be true for all-natural numbers using the principle of mathematical induction.

Complete step-by-step solution:
According to the question, we need to prove that ${{x}^{n}}-{{y}^{n}}$ is divisible by $x-y$ for all-natural numbers. Now, let us take $P\left( n \right)={{x}^{n}}-{{y}^{n}}$ is divisible by $x-y$.
Now, we will prove the given question using the principle of mathematical induction.
Therefore, first let n=1 and then we get $P\left( 1 \right)={{x}^{1}}-{{y}^{1}}$ and this is clearly divisible by $x-y$.
For n=k, we have $P\left( k \right)={{x}^{k}}-{{y}^{k}}$ and let us assume that it is divisible by $x-y$, that is the result is true for n=k.
This implies that $\dfrac{{{x}^{k}}-{{y}^{k}}}{x-y}=c$ where c is any non-zero constant.
\begin{align} & \Rightarrow {{x}^{k}}-{{y}^{k}}=c\left( x-y \right) \\ & \Rightarrow {{x}^{k}}={{y}^{k}}+c\left( x-y \right) \\ \end{align}
Now, let us take n=k+1, then we get, $P\left( k+1 \right)={{x}^{k+1}}-{{y}^{k+1}}$ now, we need to show that this is also divisible by $x-y$.
Now, we can write $P\left( k+1 \right)={{x}^{k+1}}-{{y}^{k+1}}$ as $P\left( k+1 \right)={{x}^{k}}x-{{y}^{k}}y$ and substituting the value of ${{x}^{k}}$ from n=k result we get,
\begin{align} & P\left( k+1 \right)=\left( {{y}^{k}}+c\left( x-y \right) \right)x-{{y}^{k}}y \\ & \Rightarrow P\left( k+1 \right)=x{{y}^{k}}+cx\left( x-y \right)-{{y}^{k}}y \\ & \Rightarrow P\left( k+1 \right)=\left( x-y \right){{y}^{k}}+cx\left( x-y \right) \\ \end{align}
Now, we can clearly see that $x-y$ is common in both the terms therefore, taking it out as common we get $P\left( k+1 \right)=\left( x-y \right)\left( {{y}^{k}}+cx \right)$ .
Now as $x-y$ is the factor of P(k+1), therefore, we can say that P(k+1) is also divisible by $x-y$.
Hence, we can say that by principal of mathematical induction that the given $P\left( n \right)={{x}^{n}}-{{y}^{n}}$ is divisible by $x-y$ for all-natural numbers.

Note: In such questions, we need to be aware of the principle of mathematical induction and from this we can also say that the given statement is true for all odd numbers and even numbers also as all natural numbers satisfy this.