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Prove that \[{x^{2n}} - {y^{2n}}\] is divisible by \[x + y\]

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Hint: Here we are given a problem in which we need to check if one term is divisible by the other one or not. We will use the help of the process of mathematical induction and check if we can prove the result. First we need to check if the expression is true for \[n = 1\], then we need to consider that it is true for \[n = k\;\]and then need to check if it’s true for \[n = k + 1\], if true for \[n = k + 1\] we can say that the equations holds true.

Complete step-by-step answer:
Let us take, \[P\left( n \right):{x^{2n}} - {y^{2n}} = \left( {x + y} \right) \times d\;\] where \[d \in \mathbb{N}\], the set of natural numbers.
We will try to prove this with the technique of mathematical induction process,
For \[n = 1\],
So, now substituting the value of n, we get,
 LHS \[ = {x^{2 \times 1}} - {y^{2 \times 1}}\]
\[ = {x^2} - {y^2}\]
\[ = \left( {x + y} \right)\left( {x - y} \right)\]
Which is clearly divisible by \[(x + y)\],
So, we get,
∴ \[P\left( n \right)\;\] is true for \[n = 1\],
Now, let us assume,
 \[P\left( k \right)\;\]is true.
So, substituting the value of n as k, we get,
\[{x^{2k}} - {y^{2k}}\] \[ = \left( {x + y} \right) \times m\;\] where \[m \in \mathbb{N}\],
Now, our task is to prove this result for \[k + 1\],
We will prove that \[P\left( {k + 1} \right)\;\]is true.
So, now,
LHS, \[ = {x^{2 \times \left( {k + 1} \right)}} - {y^{2 \times (k + 1)}}\]
\[ = {x^{2k + 2}} - {y^{2k + 2}}\]
On using \[{{\text{x}}^{a + b}} = {x^a}.{x^b}\], we get,
\[ = {x^{2k}}.{x^2} - {y^{2k}}.{y^2}\]
Now, as \[P\left( k \right)\;\]is true, we have,
\[{x^{2k}} - {y^{2k}} = \left( {x + y} \right) \times m\;\]
\[ \Rightarrow {x^{2k}} - {y^{2k}} = mx + my\]
\[ \Rightarrow {x^{2k}} = {y^{2k}} + mx + my\]
Substituting the value of \[{x^{2k}}\]in \[{x^{2k}}.{x^2} - {y^{2k}}.{y^2}\], we get,
\[ = ({y^{2k}} + m(x + y)).{x^2} - {y^{2k}}.{y^2}\]
On simplifying we get,
\[ = {x^2}{y^{2k}} + {x^2}m(x + y) - {y^{2k}}.{y^2}\]
On taking \[{y^{2k}}\] common we get,
\[ = ({x^2} - {y^2}){y^{2k}} + {x^2}m(x + y)\]
Taking \[(x + y)\]common, we get,
\[ = (x + y)[(x - y){y^{2k}} + {x^2}m]\]
\[ = \left( {x + y} \right) \times r\]
where, \[r = \left[ {m{x^2} + {y^{2k}}\left( {x - y} \right)} \right]\]
So, now we get, \[P\left( {k + 1} \right)\;\] is true whenever \[P\left( k \right)\;\] is true.
As, p(n) is true for \[n = 1\], \[n = k\;\]and \[n = k + 1\], hence we can say that \[{\text{p(n)}}\] is true\[\forall n \in N\].
Hence, proved.

Note: Mathematical induction is a mathematical proof technique. It is essentially used to prove that a statement \[P\left( n \right)\]holds for every natural number \[n\; = \;0,\;1,\;2,\;3,\;\]. . . ; that is, the overall statement is a sequence of infinitely many cases \[P\left( 0 \right),\;P\left( 1 \right),\;P\left( 2 \right),\;P\left( 3 \right),\]. . . . . We generally assume that \[P\left( n \right)\]is true for \[n = k\;\]and using this we prove that \[P\left( n \right)\] is true for \[n = k + 1\].