Find All Complex Solutions (x^3+9)(x^2-4)=0

$(x^{3} + 9) (x^{2} - 4) = 0$

Step 1

Factor the left side of the equation.

Rewrite $4$ as $2^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{3} + 9 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 3

Set $x^{3} + 9$ equal to $0$ and solve for $x$ .

Set $x^{3} + 9$ equal to $0$ .

$x^{3} + 9 = 0$

Solve $x^{3} + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$x^{3} = - 9$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{- 9}$

Simplify $\sqrt[3]{- 9}$ .

Rewrite $- 9$ as ${(- 1)}^{3} \cdot 9$ .

Rewrite $- 9$ as $- 1 (9)$ .

$x = \sqrt[3]{- 1 (9)}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$x = \sqrt[3]{{(- 1)}^{3} \cdot 9}$

Pull terms out from under the radical.

$x = - 1 \sqrt[3]{9}$

Rewrite $- 1 \sqrt[3]{9}$ as $- \sqrt[3]{9}$ .

$x = - \sqrt[3]{9}$

Step 4

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 6

The final solution is all the values that make $(x^{3} + 9) (x + 2) (x - 2) = 0$ true.

$x = - \sqrt[3]{9}, - 2, 2$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - \sqrt[3]{9}, - 2, 2$

Decimal Form:

$x = - 2.08008382 \dots, - 2, 2$

Find All Complex Solutions (x^3+9)(x^2-4)=0

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