Solve for a 3a^3-10=-7

$3 a^{3} - 10 = - 7$

Move all terms not containing $a$ to the right side of the equation.

Add $10$ to both sides of the equation.

$3 a^{3} = - 7 + 10$

Add $- 7$ and $10$ .

$3 a^{3} = 3$

Subtract $3$ from both sides of the equation.

$3 a^{3} - 3 = 0$

Factor the left side of the equation.

Factor $3$ out of $3 a^{3} - 3$ .

Factor $3$ out of $3 a^{3}$ .

$3 (a^{3}) - 3 = 0$

Factor $3$ out of $- 3$ .

$3 (a^{3}) + 3 (- 1) = 0$

Factor $3$ out of $3 (a^{3}) + 3 (- 1)$ .

$3 (a^{3} - 1) = 0$

Rewrite $1$ as $1^{3}$ .

$3 (a^{3} - 1^{3}) = 0$

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 $a = a$ and $b = 1$ .

$3 ((a - 1) (a^{2} + a \cdot 1 + 1^{2})) = 0$

Factor.

Simplify.

Multiply $a$ by $1$ .

$3 ((a - 1) (a^{2} + a + 1^{2})) = 0$

One to any power is one.

$3 ((a - 1) (a^{2} + a + 1)) = 0$

Remove unnecessary parentheses.

$3 (a - 1) (a^{2} + a + 1) = 0$

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

$a - 1 = 0$

$a^{2} + a + 1 = 0$

Set $a - 1$ equal to $0$ and solve for $a$ .

Set $a - 1$ equal to $0$ .

$a - 1 = 0$

Add $1$ to both sides of the equation.

$a = 1$

Set $a^{2} + a + 1$ equal to $0$ and solve for $a$ .

Set $a^{2} + a + 1$ equal to $0$ .

$a^{2} + a + 1 = 0$

Solve $a^{2} + a + 1 = 0$ for $a$ .

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

$a = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$a = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

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

$a = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

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

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

Multiply $2$ by $1$ .

$a = \frac{- 1 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$a = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

The final solution is all the values that make $3 (a - 1) (a^{2} + a + 1) = 0$ true.

$a = 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Solve for a 3a^3-10=-7

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