Question

How do you factor $x^5-y^5$?

Answer 1

Factorize using the algebraic identity:

The given expression is $x^5-y^5$.

According to the algebraic identity,

${\mathrm{a}}^{\mathrm{n}}\mathrm{-}{\mathrm{b}}^{\mathrm{n}}\mathrm{=}\mathrm{(}\mathrm{a}\mathrm{-}\mathrm{b}\mathrm{)}\mathrm{(}{\mathrm{a}}^{\mathrm{n}\mathrm{-}\mathrm{1}}\mathrm{+}{\mathrm{a}}^{\mathrm{n}\mathrm{-}\mathrm{2}}\mathrm{b}\mathrm{+}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{+}{\mathrm{ab}}^{\mathrm{n}\mathrm{-}\mathrm{2}}\mathrm{+}{\mathrm{b}}^{\mathrm{n}\mathrm{-}\mathrm{1}}\mathrm{)}$

So, the given expression can be factorized as follows,

$x^5-y^5=\left(x-y\right)\left(x^{5-1}+x^{5-2}y+x^{5-3}{y}^2+x^{5-4}{y}^3+x^{5-5}{y}^4\right)$

$\Rightarrow x^5-y^5=\left(x-y\right)\left(x^4+x^{3}y+x^2{y}^2+x^1{y}^3+x^0{y}^4\right)$

$\Rightarrow x^5-y^5=\left(x-y\right)\left(x^4+x^{3}y+x^2{y}^2+x{y}^3+1\times y^4\right)$ $\mathrm{[}\mathrm{\because }\mathrm{}{\mathrm{a}}^{\mathrm{0}}\mathrm{=}\mathrm{1}\mathrm{]}$

$\Rightarrow x^5-y^5=\left(x-y\right)\left(x^4+x^{3}y+x^2{y}^2+x{y}^3+y^4\right)$

Hence, the factors of $x^5-y^5$ obtained by using the algebraic identity is $\left(x-y\right)\left(x^4+x^{3}y+x^2{y}^2+x{y}^3+y^4\right)$.