Question: Solve for $u$, where $u$ is a real number.
$$\sqrt[3]{2u - 8} = \sqrt[3]{7u + 2}$$
If there is more than one solution, separate them with commas.
If there is no solution, click on "No solution".
$u =$
1. **State the problem:** We need to solve for the real number $u$ such that $$\sqrt[3]{2u - 8} = \sqrt[3]{7u + 2}.$$
2. **Use the property of cube roots:** Since cube root is a one-to-one function for real numbers, if $$\sqrt[3]{a} = \sqrt[3]{b},$$ then $$a = b.$$
3. **Set the insides equal:**
$$2u - 8 = 7u + 2.$$
4. **Solve the linear equation:**
Subtract $7u$ from both sides:
$$2u - 7u - 8 = 2$$
$$\cancel{2u} - 7u - 8 = 2$$
$$-5u - 8 = 2.$$
Add $8$ to both sides:
$$-5u - 8 + 8 = 2 + 8$$
$$-5u = 10.$$
Divide both sides by $-5$:
$$\frac{-5u}{\cancel{-5}} = \frac{10}{\cancel{-5}}$$
$$u = -2.$$
5. **Check the solution:** Substitute $u = -2$ back into the original equation:
$$\sqrt[3]{2(-2) - 8} = \sqrt[3]{7(-2) + 2}$$
$$\sqrt[3]{-4 - 8} = \sqrt[3]{-14 + 2}$$
$$\sqrt[3]{-12} = \sqrt[3]{-12}$$
Both sides are equal, so $u = -2$ is a valid solution.
**Final answer:**
$$u = -2.$$