Question

How do you factor completely $x^3+y^3+z^3-3xyz$?

Answer 1

Factorize the given expression using algebraic identity

The given expression is $x^3+y^3+z^3-3xyz$.

It can be factorized as follows,

$x^3+y^3+z^3-3xyz=x^3+y^3+z^3-3xyz+3x^{2}y-3x^{2}y+3x{y}^2-3x{y}^2$

$\Rightarrow x^3+y^3+z^3-3xyz=\left[x^3+y^3+3x^{2}y+3x{y}^2\right]+z^3-\left[3xyz+3x^{2}y+3x{y}^2\right]$

$\Rightarrow x^3+y^3+z^3-3xyz=\left[x^3+y^3+3xy\left(x+y\right)\right]+z^3-3xy\left[z+x+y\right]$

$\Rightarrow x^3+y^3+z^3-3xyz=\left[{\left(x+y\right)}^3\right]+z^3-3xy\left(x+y+z\right)$ $\left[\because {\left(x+y\right)}^3=x^3+y^3+3xy\left(x+y\right)\right]$

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

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

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

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

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

$\Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)$

$\Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left[\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\right]$ $\left[\because ab+ac=a\left(b+c\right)\right]$

$\Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2-3xy\right)$

$\Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$

Hence, the factors of $x^3+y^3+z^3-3xyz$ obtained by using algebraic identity is $\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$.