Prove by induction
\[{{x}^{n}}-{{y}^{n}}\] is divisible by (x+y) when n is even.

HINT: Mathematical induction is a mathematical proof technique. A proof by induction consists of two cases. The first, the base case (or basis), proves the statement for n = 0 without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case n = k, then it must also hold for the next case n = k + 1. These two steps establish that the statement holds for every natural number n.[3] The base case does not necessarily begin with n = 0, but often with n = 1, and possibly with any fixed natural number n = N, establishing the truth of the statement for all natural numbers n ≥ N.

Complete step-by-step answer:
As mentioned in the question, we have to prove the statement through mathematical induction.
As mentioned in the hint,
Let P(n) \[={{x}^{n}}-{{y}^{n}}\] is divisible by (x+y)
Case 1:-
For n=2(being the smallest even number), we get
P(2)
\[\begin{align}
  & ={{x}^{2}}-{{y}^{2}} \\
 & =(x+y)(x-y) \\
\end{align}\]
Hence, P(n) is true for n=2.
Case 2:-
For n=2k(an even number)
Let P(k) ‘ \[={{x}^{2k}}-{{y}^{2k}}\] is divisible by x+y ‘ be true.
Hence, we can write
\[\begin{align}
  & {{x}^{2k}}-{{y}^{2k}}=n(x+y) \\
 & {{x}^{2k}}=\ n(x+y)+\ {{y}^{2k}}\ \ \ ...(a) \\
\end{align}\]
Case 3:-
Now, we have to prove P(k+1) to be true.
Hence,
 P(k+1)
\[\begin{align}
  & ={{x}^{2(k+1)}}-{{y}^{2(k+1)}} \\
 & ={{x}^{2k+2}}-{{y}^{2k+2}} \\
 & ={{x}^{2k}}\cdot {{x}^{2}}-{{y}^{2k}}.{{y}^{2}} \\
\end{align}\]
Now, using equation (a), we get
\[\begin{align}
  & ={{x}^{2k}}\cdot {{x}^{2}}-{{y}^{2k}}.{{y}^{2}} \\
 & =\left( n(x+y)+\ {{y}^{2k}} \right)\cdot {{x}^{2}}-{{y}^{2k}}.{{y}^{2}} \\
 & =n(x+y)\cdot {{x}^{2}}+{{y}^{2k}}({{x}^{2}}-{{y}^{2}}) \\
 & =n(x+y)\cdot {{x}^{2}}+{{y}^{2k}}(x-y)(x+y) \\
 & =(x+y)\left( n{{x}^{2}}+{{y}^{2k}}(x-y) \right)\ \ \ \ \ ...(b) \\
\end{align}\]
As equation (b) is divisible by (x+y), therefore, P(k+1) is true .
Hence, the statement is proved by use of mathematical induction.
     \[\]

NOTE: A student can make an error if they don’t know about the process of mathematical induction which is as follows
A proof by induction consists of two cases. The first, the base case (or basis), proves the statement for n = 0 without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case n = k, then it must also hold for the next case n = k + 1. These two steps establish that the statement holds for every natural number n.[3] The base case does not necessarily begin with n = 0, but often with n = 1, and possibly with any fixed natural number n = N, establishing the truth of the statement for all natural numbers n ≥ N.