1. **State the problem:** Last year, there were $z$ zebra mussels in a river section. This year, there are $z^3$ zebra mussels, and this year the number is 729. Find $z$. 2. **Write the equation:** Since this year there are $z^3$ mussels and that equals 729, we have: $$z^3 = 729$$ 3. **Solve for $z$:** To find $z$, take the cube root of both sides: $$z = \sqrt[3]{729}$$ 4. **Calculate the cube root:** Since $9 \times 9 \times 9 = 729$, we get: $$z = 9$$ 5. **Answer:** There were 9 zebra mussels last year. --- 1. **State the problem:** Solve the equation: $$y^3 + 15 = 140$$ 2. **Isolate $y^3$:** Subtract 15 from both sides: $$y^3 + 15 - 15 = 140 - 15$$ $$y^3 = 125$$ 3. **Take cube root:** $$y = \sqrt[3]{125}$$ 4. **Calculate cube root:** Since $5 \times 5 \times 5 = 125$, $$y = 5$$ 5. **Answer:** The solution is $y = 5$. --- 1. **Complete the following:** - $\sqrt{0} = 0$ because the square root of zero is zero. - $\sqrt[3]{0} = 0$ because the cube root of zero is zero. - $\sqrt{1} = 1$ because the square root of one is one. - $\sqrt[3]{1} = 1$ because the cube root of one is one.