Question

How do you factor ${\mathrm{x}}^3+{\mathrm{y}}^3$?

Answer 1

Factorize the given expression:

Given expression is ${\mathrm{x}}^3+{\mathrm{y}}^3$

We know that ${\left(\mathrm{x}+\mathrm{y}\right)}^3={\mathrm{x}}^3+{\mathrm{y}}^3+3\mathrm{xy}\left(\mathrm{x}+\mathrm{y}\right)$

$\Rightarrow$ ${\left(\mathrm{x}+\mathrm{y}\right)}^3-3\mathrm{xy}\left(\mathrm{x}+\mathrm{y}\right)={\mathrm{x}}^3+{\mathrm{y}}^3$

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

$\Rightarrow$ $\left(\mathrm{x}+\mathrm{y}\right)\left[{\left(\mathrm{x}+\mathrm{y}\right)}^2-3\mathrm{xy}\right]={\mathrm{x}}^3+{\mathrm{y}}^3$

$\Rightarrow$$\left(\mathrm{x}+\mathrm{y}\right)\left[{\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xy}-3\mathrm{xy}\right]={\mathrm{x}}^3+{\mathrm{y}}^3\left[\because {\left(\mathrm{a}+\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\right]$

$\Rightarrow$ $\left(\mathrm{x}+\mathrm{y}\right)\left[{\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{xy}\right]={\mathrm{x}}^3+{\mathrm{y}}^3$

$\Rightarrow$ ${\mathrm{x}}^3+{\mathrm{y}}^3=\left(\mathrm{x}+\mathrm{y}\right)\left[{\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{xy}\right]$

Hence, factorized form of ${\mathrm{x}}^3+{\mathrm{y}}^3$is $\left(\mathrm{x}+\mathrm{y}\right)\left[{\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{xy}\right]$