# How do you solve x^3-3x^2-x+3=0?

##### 1 Answer

The sum of all coefficients of given cubic equation: ${x}^{3} - 3 {x}^{2} - x + 3 = 0$ is zero hence $x = 1$ is a root of given cubic equation i.e. $\left(x - 1\right)$ is a factor of ${x}^{3} - 3 {x}^{3} - x + 3$
Now, cubic algebraic polynomial: ${x}^{3} - 3 {x}^{2} - x + 3$ can be factorized as follows
${x}^{3} - 3 {x}^{3} - x + 3$
$= {x}^{2} \left(x - 1\right) - 2 x \left(x - 1\right) - 3 \left(x - 1\right)$
$= \left(x - 1\right) \left({x}^{2} - 2 x - 3\right)$
$= \left(x - 1\right) \left({x}^{2} - 3 x + x - 3\right)$
$= \left(x - 1\right) \left(x \left(x - 3\right) + \left(x - 3\right)\right)$
$= \left(x - 1\right) \left(x - 3\right) \left(x + 1\right)$
hence, the solution of given cubic equation will be given as
$\left(x - 1\right) \left(x - 3\right) \left(x + 1\right) = 0$
$x = - 1 , 1 , 3$