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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.