Lesson 10: Practice Problems

Problem 1

What is the volume of a cube with a side length of
- 4 centimeters?
- $\sqrt[3]{11}$ feet?
- $s$ units?
What is the side length of a cube with a volume of
- 1,000 cubic centimeters?
- 23 cubic inches?
- $v$ cubic units?

Problem 2

Write an equivalent expression that doesn’t use a cube root symbol.

$\sqrt[3]{1}$
$\sqrt[3]{216}$
$\sqrt[3]{8000}$
$\sqrt[3]{\frac{1}{64}}$
$\sqrt[3]{\frac{27}{125}}$
$\sqrt[3]{0.027}$
$\sqrt[3]{0.000125}$

Problem 3

Find the positive solution to each equation. If the solution is irrational, write the solution using square root or cube root notation.

$t^{3} = 216$
$a^{2} = 15$
$m^{3} = 8$
$c^{3} = 343$
$f^{3} = 181$

Problem 4

For each cube root, find the two whole numbers that it lies between.

$\sqrt[3]{11}$
$\sqrt[3]{80}$
$\sqrt[3]{120}$
$\sqrt[3]{250}$

Problem 5

Order the following values from least to greatest:

$\sqrt[3]{530}$
$\sqrt{48}$
$π$
$\sqrt{121}$
$\sqrt[3]{27}$
$\frac{19}{2}$

Problem 6

The equation $x^{2} = 25$ has two solutions. This is because both $5 \cdot 5 = 25$ , and also $- 5 \cdot - 5 = 25$ . So, 5 is a solution, and also -5 is a solution. But! The equation $x^{3} = 125$ only has one solution, which is 5. This is because $5 \cdot 5 \cdot 5 = 125$ , and there are no other numbers you can cube to make 125. (Think about why -5 is not a solution!)

Find all the solutions to each equation.

$x^{3} = 8$
$\sqrt[3]{x} = 3$
$x^{2} = 49$
$x^{3} = \frac{64}{125}$