Solve for h 2h^3-11=43

$2 h^{3} - 11 = 43$

Step 1

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

Add $11$ to both sides of the equation.

$2 h^{3} = 43 + 11$

Add $43$ and $11$ .

$2 h^{3} = 54$

Step 2

Subtract $54$ from both sides of the equation.

$2 h^{3} - 54 = 0$

Step 3

Factor the left side of the equation.

Factor $2$ out of $2 h^{3} - 54$ .

Factor $2$ out of $2 h^{3}$ .

$2 (h^{3}) - 54 = 0$

Factor $2$ out of $- 54$ .

$2 h^{3} + 2 \cdot - 27 = 0$

Factor $2$ out of $2 h^{3} + 2 \cdot - 27$ .

$2 (h^{3} - 27) = 0$

Rewrite $27$ as $3^{3}$ .

$2 (h^{3} - 3^{3}) = 0$

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 $a = h$ and $b = 3$ .

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

Factor.

Simplify.

Move $3$ to the left of $h$ .

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

Raise $3$ to the power of $2$ .

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

Remove unnecessary parentheses.

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

Step 4

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

$h - 3 = 0$

$h^{2} + 3 h + 9 = 0$

Step 5

Set $h - 3$ equal to $0$ and solve for $h$ .

Set $h - 3$ equal to $0$ .

$h - 3 = 0$

Add $3$ to both sides of the equation.

$h = 3$

Step 6

Set $h^{2} + 3 h + 9$ equal to $0$ and solve for $h$ .

Set $h^{2} + 3 h + 9$ equal to $0$ .

$h^{2} + 3 h + 9 = 0$

Solve $h^{2} + 3 h + 9 = 0$ for $h$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 3$ , and $c = 9$ into the quadratic formula and solve for $h$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$h = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 9$ .

Multiply $- 4$ by $1$ .

$h = \frac{- 3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$h = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$h = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$h = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$h = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$h = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$h = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical.

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

Move $3$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

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

Step 7

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

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

Solve for h 2h^3-11=43

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