Solve for d d^3=512

$d^{3} = 512$

Subtract $512$ from both sides of the equation.

$d^{3} - 512 = 0$

Factor the left side of the equation.

Rewrite $512$ as $8^{3}$ .

$d^{3} - 8^{3} = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = d$ and $b = 8$ .

$(d - 8) (d^{2} + d \cdot 8 + 8^{2}) = 0$

Simplify.

Move $8$ to the left of $d$ .

$(d - 8) (d^{2} + 8 d + 8^{2}) = 0$

Raise $8$ to the power of $2$ .

$(d - 8) (d^{2} + 8 d + 64) = 0$

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

$d - 8 = 0$

$d^{2} + 8 d + 64 = 0$

Set $d - 8$ equal to $0$ and solve for $d$ .

Set $d - 8$ equal to $0$ .

$d - 8 = 0$

Add $8$ to both sides of the equation.

$d = 8$

Set $d^{2} + 8 d + 64$ equal to $0$ and solve for $d$ .

Set $d^{2} + 8 d + 64$ equal to $0$ .

$d^{2} + 8 d + 64 = 0$

Solve $d^{2} + 8 d + 64 = 0$ for $d$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 8$ , and $c = 64$ into the quadratic formula and solve for $d$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (1 \cdot 64)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$d = \frac{- 8 \pm \sqrt{64 - 4 \cdot 1 \cdot 64}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$d = \frac{- 8 \pm \sqrt{64 - 4 \cdot 64}}{2 \cdot 1}$

Multiply $- 4$ by $64$ .

$d = \frac{- 8 \pm \sqrt{64 - 256}}{2 \cdot 1}$

Subtract $256$ from $64$ .

$d = \frac{- 8 \pm \sqrt{- 192}}{2 \cdot 1}$

Rewrite $- 192$ as $- 1 (192)$ .

$d = \frac{- 8 \pm \sqrt{- 1 \cdot 192}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (192)}$ as $\sqrt{- 1} \cdot \sqrt{192}$ .

$d = \frac{- 8 \pm \sqrt{- 1} \cdot \sqrt{192}}{2 \cdot 1}$

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

$d = \frac{- 8 \pm i \cdot \sqrt{192}}{2 \cdot 1}$

Rewrite $192$ as $8^{2} \cdot 3$ .

Factor $64$ out of $192$ .

$d = \frac{- 8 \pm i \cdot \sqrt{64 (3)}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$d = \frac{- 8 \pm i \cdot \sqrt{8^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$d = \frac{- 8 \pm i \cdot (8 \sqrt{3})}{2 \cdot 1}$

Move $8$ to the left of $i$ .

$d = \frac{- 8 \pm 8 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$d = \frac{- 8 \pm 8 i \sqrt{3}}{2}$

Simplify $\frac{- 8 \pm 8 i \sqrt{3}}{2}$ .

$d = - 4 \pm 4 i \sqrt{3}$

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(d - 8) (d^{2} + 8 d + 64) = 0$ true.

$d = 8, - 4 + 4 i \sqrt{3}, - 4 - 4 i \sqrt{3}$

Solve for d d^3=512

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