Question

Simplify: $x^6-y^6$

Answer 1

Consider $x^6-y^6$

${\left(x^3\right)}^2-{\left(y^3\right)}^2$

We know that $\left(a+b\right)\left(a-b\right)=a^2-b^2$

Here $a=x^3,b=y^3$

$$\begin{gathered} \therefore {\left(x^3\right)}^2-{\left(y^3\right)}^2 \\ \left(x^3-y^3\right)\left(x^3+y^3\right) \end{gathered}$$

Also we know that $$\begin{gathered} a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right) \\ {a}^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right) \end{gathered}$$

$$\begin{gathered} \left(x^3-y^3\right)\left(x^3+y^3\right) \\ \left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right) \\ \left(x+y\right)\left(x-y\right)\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right) \\ \left(x+y\right)\left(x-y\right)[x^2\left(x^2+xy+y^2\right)-xy\left(x^2+xy+y^2\right)+y^2\left(x^2+xy+y^2\right)] \\ \left(x+y\right)\left(x-y\right)\left(x^4+x^{3}y+x^2{y}^2-x^{3}y-x^2{y}^2-x{y}^3+y^2{x}^2+x{y}^3+y^3\right) \\ \left(x+y\right)\left(x-y\right)\left(x^4+y^2{x}^2+y^4\right) \end{gathered}$$

Hence, $x^6-y^6$ is $\left(x+y\right)\left(x-y\right)\left(x^4+x^2{y}^2+y^3\right)$