We already know that $x^2 - y^2 = (x - y)(x + y)$.
 
Further, we have verified that $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$.

Observe that $x - y$ is a common factor of both $x^2 - y^2$ and $x^3 - y^3$.

Do you think $x - y$ also a factor of $x^4 - y^4$? Justify your answer.

Note that $x^4 - y^4 = (x^2)^2 - (y^2)^2$.

Can you see how $x - y$ is a factor of $x^4 - y^4$?

How about $x^5 - y^5$? Does this also have $x - y$ as a factor?                          [Page No. 85]

Yes, in both cases $x - y$ is a factor.

For $x^4 - y^4$,

$$x^4 - y^4 = (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2).$$

Hence, $x - y$ is a factor of $x^4 - y^4$.

For $x^5 - y^5$,

$$x^5 - y^5 = (x - y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4).$$

Thus, $x - y$ is also a factor of $x^5 - y^5$.

In fact, this leads to a general result:

$$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$

for every positive integer $n$.

Therefore, $x - y$ is always a factor of $x^n - y^n$.