Solve for t t^3=64/27

$t^{3} = \frac{64}{27}$

Step 1

Subtract $\frac{64}{27}$ from both sides of the equation.

$t^{3} - \frac{64}{27} = 0$

Step 2

Factor the left side of the equation.

Rewrite $\frac{64}{27}$ as ${(\frac{4}{3})}^{3}$ .

$t^{3} - {(\frac{4}{3})}^{3} = 0$

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 $a = t$ and $b = \frac{4}{3}$ .

$(t - \frac{4}{3}) (t^{2} + t (\frac{4}{3}) + {(\frac{4}{3})}^{2}) = 0$

Simplify.

Combine $t$ and $\frac{4}{3}$ .

$(t - \frac{4}{3}) (t^{2} + \frac{t \cdot 4}{3} + {(\frac{4}{3})}^{2}) = 0$

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

$(t - \frac{4}{3}) (t^{2} + \frac{4 \cdot t}{3} + {(\frac{4}{3})}^{2}) = 0$

Multiply $4$ by $t$ .

$(t - \frac{4}{3}) (t^{2} + \frac{4 t}{3} + {(\frac{4}{3})}^{2}) = 0$

Apply the product rule to $\frac{4}{3}$ .

$(t - \frac{4}{3}) (t^{2} + \frac{4 t}{3} + \frac{4^{2}}{3^{2}}) = 0$

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

$(t - \frac{4}{3}) (t^{2} + \frac{4 t}{3} + \frac{16}{3^{2}}) = 0$

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

$(t - \frac{4}{3}) (t^{2} + \frac{4 t}{3} + \frac{16}{9}) = 0$

Step 3

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

$t - \frac{4}{3} = 0$

$t^{2} + \frac{4 t}{3} + \frac{16}{9} = 0$

Step 4

Set $t - \frac{4}{3}$ equal to $0$ and solve for $t$ .

Set $t - \frac{4}{3}$ equal to $0$ .

$t - \frac{4}{3} = 0$

Add $\frac{4}{3}$ to both sides of the equation.

$t = \frac{4}{3}$

Step 5

Set $t^{2} + \frac{4 t}{3} + \frac{16}{9}$ equal to $0$ and solve for $t$ .

Set $t^{2} + \frac{4 t}{3} + \frac{16}{9}$ equal to $0$ .

$t^{2} + \frac{4 t}{3} + \frac{16}{9} = 0$

Solve $t^{2} + \frac{4 t}{3} + \frac{16}{9} = 0$ for $t$ .

Multiply through by the least common denominator $9$ , then simplify.

Apply the distributive property.

$9 t^{2} + 9 (\frac{4 t}{3}) + 9 (\frac{16}{9}) = 0$

Simplify.

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$9 t^{2} + 3 (3) (\frac{4 t}{3}) + 9 (\frac{16}{9}) = 0$

Cancel the common factor.

$9 t^{2} + 3 \cdot (3 (\frac{4 t}{3})) + 9 (\frac{16}{9}) = 0$

Rewrite the expression.

$9 t^{2} + 3 (4 t) + 9 (\frac{16}{9}) = 0$

Multiply $4$ by $3$ .

$9 t^{2} + 12 t + 9 (\frac{16}{9}) = 0$

Cancel the common factor of $9$ .

Cancel the common factor.

$9 t^{2} + 12 t + 9 (\frac{16}{9}) = 0$

Rewrite the expression.

$9 t^{2} + 12 t + 16 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 9$ , $b = 12$ , and $c = 16$ into the quadratic formula and solve for $t$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (9 \cdot 16)}}{2 \cdot 9}$

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$t = \frac{- 12 \pm \sqrt{144 - 4 \cdot 9 \cdot 16}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot 16$ .

Multiply $- 4$ by $9$ .

$t = \frac{- 12 \pm \sqrt{144 - 36 \cdot 16}}{2 \cdot 9}$

Multiply $- 36$ by $16$ .

$t = \frac{- 12 \pm \sqrt{144 - 576}}{2 \cdot 9}$

Subtract $576$ from $144$ .

$t = \frac{- 12 \pm \sqrt{- 432}}{2 \cdot 9}$

Rewrite $- 432$ as $- 1 (432)$ .

$t = \frac{- 12 \pm \sqrt{- 1 \cdot 432}}{2 \cdot 9}$

Rewrite $\sqrt{- 1 (432)}$ as $\sqrt{- 1} \cdot \sqrt{432}$ .

$t = \frac{- 12 \pm \sqrt{- 1} \cdot \sqrt{432}}{2 \cdot 9}$

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

$t = \frac{- 12 \pm i \cdot \sqrt{432}}{2 \cdot 9}$

Rewrite $432$ as $12^{2} \cdot 3$ .

Factor $144$ out of $432$ .

$t = \frac{- 12 \pm i \cdot \sqrt{144 (3)}}{2 \cdot 9}$

Rewrite $144$ as $12^{2}$ .

$t = \frac{- 12 \pm i \cdot \sqrt{12^{2} \cdot 3}}{2 \cdot 9}$

Pull terms out from under the radical.

$t = \frac{- 12 \pm i \cdot (12 \sqrt{3})}{2 \cdot 9}$

Move $12$ to the left of $i$ .

$t = \frac{- 12 \pm 12 i \sqrt{3}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$t = \frac{- 12 \pm 12 i \sqrt{3}}{18}$

Simplify $\frac{- 12 \pm 12 i \sqrt{3}}{18}$ .

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

The final answer is the combination of both solutions.

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

Step 6

The final solution is all the values that make $(t - \frac{4}{3}) (t^{2} + \frac{4 t}{3} + \frac{16}{9}) = 0$ true.

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

Solve for t t^3=64/27

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