Question: Solve $(v-2)^3 + 64 = 0$ where $v$ is a real number.
Write your answer in simplified radical form.
(If there is more than one solution, separate them with commas.)
$v = $
1. **State the problem:** Solve the equation $$ (v-2)^3 + 64 = 0 $$ for real $v$.
2. **Rewrite the equation:** Recognize that $$64 = 4^3$$, so the equation becomes $$ (v-2)^3 + 4^3 = 0 $$.
3. **Use the sum of cubes formula:** Recall that $$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$.
4. **Apply the formula:** Let $$a = v-2$$ and $$b = 4$$, then
$$ (v-2)^3 + 4^3 = (v-2 + 4)((v-2)^2 - (v-2)(4) + 4^2) = 0 $$.
5. **Simplify the factors:**
- First factor: $$v - 2 + 4 = v + 2$$
- Second factor:
$$ (v-2)^2 - 4(v-2) + 16 = (v^2 - 4v + 4) - 4v + 8 + 16 = v^2 - 8v + 28 $$
6. **Set each factor equal to zero:**
- $$v + 2 = 0 \\ \Rightarrow v = -2$$
- $$v^2 - 8v + 28 = 0$$
7. **Solve the quadratic:**
Use the quadratic formula:
$$ v = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 28}}{2} = \frac{8 \pm \sqrt{64 - 112}}{2} = \frac{8 \pm \sqrt{-48}}{2} $$
8. **Simplify the discriminant:**
$$ \sqrt{-48} = \sqrt{-1 \cdot 16 \cdot 3} = 4i\sqrt{3} $$
9. **Write the complex solutions:**
$$ v = \frac{8 \pm 4i\sqrt{3}}{2} = 4 \pm 2i\sqrt{3} $$
10. **Identify real solutions:** Only $$v = -2$$ is real.
**Final answer:**
$$ v = -2 $$