# How do you solve 3x^7 - 243x^3=0?

Dec 12, 2015

$x = 0 , 3$

#### Explanation:

First thing you want to do is factor. You can factor out $3 {x}^{3}$

$\left(3 {x}^{3}\right) \left({x}^{4} - 81\right) = 0$

Now set both factors to zero

$3 {x}^{3} = 0$

${x}^{3} = 0$

$\sqrt{{x}^{3}} = \sqrt{0}$

$x = 0$

${x}^{4} - 81 = 0$

${x}^{4} = 81$

$\sqrt{{x}^{4}} = \sqrt{81}$

$x = 3$

Dec 12, 2015

$x = 0 , \pm 3$

#### Explanation:

Factor out a common $3 {x}^{3}$.

$3 {x}^{3} \left({x}^{4} - 81\right) = 0$

Recognize a difference of squares.

$3 {x}^{3} \left({x}^{2} + 9\right) \left({x}^{2} - 9\right) = 0$

Recognize another difference of squares.

$3 {x}^{3} \left({x}^{2} + 9\right) \left(x + 3\right) \left(x - 3\right) = 0$

Set each part equal to $0$.

$3 {x}^{3} = 0$
$x = 0$

${x}^{2} + 9 = 0$
${x}^{2} = - 9$
$x = \pm 3 i$ (NONREAL ANSWERS)

$x + 3 = 0$
$x = - 3$

$x - 3 = 0$
$x = 3$

Thus, $x = 0 , \pm 3$.

Look at a graph:

graph{3x^7-243x^3 [-12.73, 12.58, -6.17, 6.48]}

The $x$-intercepts (the roots) are when $x = 0 , \pm 3$.