Algebra Examples

$q^{3} + 64 = 0$

Step 1

Subtract $64$ from both sides of the equation.

$q^{3} = - 64$

Step 2

Add $64$ to both sides of the equation.

$q^{3} + 64 = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Rewrite $64$ as $4^{3}$ .

$q^{3} + 4^{3} = 0$

Step 3.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = q$ and $b = 4$ .

$(q + 4) (q^{2} - q \cdot 4 + 4^{2}) = 0$

Step 3.3

Simplify.

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Step 3.3.1

Multiply $4$ by $- 1$ .

$(q + 4) (q^{2} - 4 q + 4^{2}) = 0$

Step 3.3.2

Raise $4$ to the power of $2$ .

$(q + 4) (q^{2} - 4 q + 16) = 0$

Step 4

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

$q + 4 = 0$

$q^{2} - 4 q + 16 = 0$

Step 5

Set $q + 4$ equal to $0$ and solve for $q$ .

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Step 5.1

Set $q + 4$ equal to $0$ .

$q + 4 = 0$

Step 5.2

Subtract $4$ from both sides of the equation.

$q = - 4$

Step 6

Set $q^{2} - 4 q + 16$ equal to $0$ and solve for $q$ .

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Step 6.1

Set $q^{2} - 4 q + 16$ equal to $0$ .

$q^{2} - 4 q + 16 = 0$

Step 6.2

Solve $q^{2} - 4 q + 16 = 0$ for $q$ .

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Step 6.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6.2.2

Substitute the values $a = 1$ , $b = - 4$ , and $c = 16$ into the quadratic formula and solve for $q$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 6.2.3

Simplify.

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Step 6.2.3.1

Simplify the numerator.

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Step 6.2.3.1.1

Raise $- 4$ to the power of $2$ .

$q = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 6.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 16$ .

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Step 6.2.3.1.2.1

Multiply $- 4$ by $1$ .

$q = \frac{4 \pm \sqrt{16 - 4 \cdot 16}}{2 \cdot 1}$

Step 6.2.3.1.2.2

Multiply $- 4$ by $16$ .

$q = \frac{4 \pm \sqrt{16 - 64}}{2 \cdot 1}$

Step 6.2.3.1.3

Subtract $64$ from $16$ .

$q = \frac{4 \pm \sqrt{- 48}}{2 \cdot 1}$

Step 6.2.3.1.4

Rewrite $- 48$ as $- 1 (48)$ .

$q = \frac{4 \pm \sqrt{- 1 \cdot 48}}{2 \cdot 1}$

Step 6.2.3.1.5

Rewrite $\sqrt{- 1 (48)}$ as $\sqrt{- 1} \cdot \sqrt{48}$ .

$q = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{48}}{2 \cdot 1}$

Step 6.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$q = \frac{4 \pm i \cdot \sqrt{48}}{2 \cdot 1}$

Step 6.2.3.1.7

Rewrite $48$ as $4^{2} \cdot 3$ .

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Step 6.2.3.1.7.1

Factor $16$ out of $48$ .

$q = \frac{4 \pm i \cdot \sqrt{16 (3)}}{2 \cdot 1}$

Step 6.2.3.1.7.2

Rewrite $16$ as $4^{2}$ .

$q = \frac{4 \pm i \cdot \sqrt{4^{2} \cdot 3}}{2 \cdot 1}$

Step 6.2.3.1.8

Pull terms out from under the radical.

$q = \frac{4 \pm i \cdot (4 \sqrt{3})}{2 \cdot 1}$

Step 6.2.3.1.9

Move $4$ to the left of $i$ .

$q = \frac{4 \pm 4 i \sqrt{3}}{2 \cdot 1}$

Step 6.2.3.2

Multiply $2$ by $1$ .

$q = \frac{4 \pm 4 i \sqrt{3}}{2}$

Step 6.2.3.3

Simplify $\frac{4 \pm 4 i \sqrt{3}}{2}$ .

$q = 2 \pm 2 i \sqrt{3}$

Step 6.2.4

The final answer is the combination of both solutions.

$q = 2 + 2 i \sqrt{3}, 2 - 2 i \sqrt{3}$

Step 7

The final solution is all the values that make $(q + 4) (q^{2} - 4 q + 16) = 0$ true.

$q = - 4, 2 + 2 i \sqrt{3}, 2 - 2 i \sqrt{3}$