# Lesson 8

Cubes and Cube Roots

## 8.1: Put Your Arm(s) Up (5 minutes)

### Warm-up

Asking students to compare $$y=x^2$$ with $$y=x^3$$ prepares students for upcoming activities in which they reason about solutions to equations involving cubes and cube roots. Just as graphs of $$y=x^2$$ and $$y=\sqrt{x}$$ helped make sense of solutions to equations involving squares and square roots, graphs of $$y=x^3$$ and $$y=\sqrt{x}$$ will help students make sense of solutions to equations involving cubes and cube roots.

### Launch

Give students a few minutes of quiet think time before a brief, whole-class discussion.

### Student Facing

How are these graphs the same? How are they different?

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### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

Ask students to share their responses for what is the same and what is different between the two graphs. Share with students that one way to remember what the graphs of $$y=x^2$$ and $$y=x^3$$ look like is to imagine holding a positive number in the right hand (the positive side of the number line) and a negative number in the left hand (the negative side of the number line). For $$y=x^2$$, squaring the positive number in the right hand produces a positive number, so the right hand goes above the shoulders. Squaring the negative number in the left hand also always produces a positive number, so the left hand also rises above the shoulders. For $$y=x^3$$, cubing the positive number in the right hand produces a positive number, so the right hand also goes above the shoulders. However, cubing a negative number in the left hand produces a negative number, so the left hand drops below the shoulders. For example, $$(\text-2)(\text-2)(\text-2)=\text- 8$$.

## 8.2: Finding Cube Roots with a Graph (10 minutes)

### Activity

In this activity, students combine what they know about graphs, solutions to equations, and cube roots to estimate solutions to cubic equations. Using the graph, students find that cubic equations of the form $$x^3=a$$ have exactly 1 solution for any value of $$a$$, positive or negative. This is different than solutions to equations involving squares and square roots students worked with in earlier lessons.

This activity works best when each student has access to the embedded applet, because the level of precision is important. If students don’t have individual access, projecting the applet would be helpful during the synthesis.

### Student Facing

Use the applet to zoom in and out on the graph.

How many solutions are there to each of the following equations? Estimate the solution(s) from the graph of $$y=x^3$$. Check your estimate by substituting it back into the equation.

1. $$x^3 = 8$$
2. $$x^3 = 2$$
3. $$x^3 = 0$$
4. $$x^3 = \text{-}8$$
5. $$x^3 = \text{-}2$$

### Student Response

For access, consult one of our IM Certified Partners.

### Student Facing

How many solutions are there to each of the following equations? Estimate the solution(s) from the graph of $$y=x^3$$. Check your estimate by substituting it back into the equation.

1. $$x^3 = 8$$
2. $$x^3 = 2$$
3. $$x^3 = 0$$
4. $$x^3 = \text{-}8$$
5. $$x^3 = \text{-}2$$

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

Select students to share a few estimates and exact solutions to the equations and explain how they used the graph to help estimate. It is important to discuss some of the differences between cube roots and square roots. Here are some questions for discussion:

• “Think back to when we were looking at the graph of $$y=x^2$$. From that graph, what did the solutions for $$x^2=a$$ look like if $$a$$ was positive, zero, or negative?” (For positive values of $$a$$, $$x^2=a$$ had two solutions, the positive square root and the negative square root. For $$x^2=0$$, there is only one solution, which is 0. For $$x^2=a$$ when $$a$$ is negative, there are no solutions because the horizontal lines below the $$x$$ axis don’t intercept the graph of $$y=x^2$$. Another perspective is that squaring a number doesn’t result in a negative number.)
• “What about the solutions to $$x^3=a$$? From the graph of $$y=x^3$$, what do the solutions look like for different values of $$a$$?” (Horizontal lines always cross the graph of $$y=x^3$$ at one place, so every number has exactly one cube root.)
Speaking: MLR8 Discussion Supports. As students share their responses to the discussion questions, press for details by asking how they know that $$x^2=a$$ has two solutions for positive values of $$a$$, one solution when $$a=0$$, and no solutions when $$a$$ is negative. Show concepts multi-modally by displaying the graph of $$y=x^2$$ and drawing the horizontal lines above and below the $$x$$ axis. This will help students justify the number of solutions for $$x^2=a$$ based on the value of $$a$$.
Design Principle(s): Support sense-making; Optimize output (for justification)

## 8.3: Cube Root Equations (10 minutes)

### Optional activity

This activity is optional because it is extra practice that not all classes may need. Students practice solving equations with a variable inside a cube root. They will have more opportunities to solve equations like these in the next activity if time is limited. Students attend to precision when they use the meaning of cube roots to find exact solutions to equations involving cube roots (MP6).

This activity works best when each student has access to the embedded applet, because the level of precision is important. If students don’t have individual access, projecting the applet would be helpful during the synthesis.

### Student Facing

Use the applet to zoom in and out on the graph.

1. Use the graph of $$y = \sqrt{x}$$ to estimate the solution(s) to $$\sqrt{x} = \text{-}4$$.
2. Use the meaning of cube roots to find an exact solution to the equation $$\sqrt{x} = \text{-}4$$. How close was your estimate?
3. Find the solution of the equation $$\sqrt{x} = 3.5$$ using the meaning of cube roots. Use the graph to check that your solution is reasonable.

### Student Response

For access, consult one of our IM Certified Partners.

### Student Facing

1. Use the graph of $$y = \sqrt{x}$$ to estimate the solution(s) to $$\sqrt{x} = \text{-}4$$.
2. Use the meaning of cube roots to find an exact solution to the equation $$\sqrt{x} = \text{-}4$$. How close was your estimate?
3. Find the solution of the equation $$\sqrt{x} = 3.5$$ using the meaning of cube roots. Use the graph to check that your solution is reasonable.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

Discuss the difference between cubing each side of $$\sqrt{x} = \text{-}4$$ to get $$x=\text- 64$$ and squaring each side of $$\sqrt{x} = \text{-}4$$ to get $$x=16$$. Ask students, “Why is one valid while the other isn’t?” (If $$\sqrt{x}$$ has a value of -4, then it cubes to make -64, and -64 has only one cube root. In contrast, the equation $$\sqrt{x} = \text{-}4$$ has no solutions because there are two numbers that square to make 16, the positive square root and negative square root, and it’s impossible for the positive square root $$\sqrt{x}$$ to be equal to the negative square root.)

## 8.4: Solve These Equations With Cube Roots in Them (20 minutes)

### Activity

In this partner activity, students take turns solving equations involving cube roots by isolating the cube root expression and then cubing each side of the equation. As students trade roles explaining their thinking and listening, they have opportunities to explain their reasoning and critique the reasoning of others (MP3).

With square roots, students would have to check that their solutions are valid by substituting in the original equation. That isn’t necessary with cube roots because there is only ever one cube root, but it is a good habit to have.

### Launch

Arrange students in groups of 2. After students have had time to identify their three most and least difficult equations, ask students to take turns: the first partner identifies an equation from one of their lists and explains why they chose it, while the other listens and works to understand. Then they switch roles. When both partners have explained why they chose one equation from their most difficult list and one from their least difficult list, they move on to the second question.

When both partners have solved the four equations they chose, ask them to take turns again: the first partner explains how they solved one of their problems, while the other listens and works to understand. When both partners agree on the solution, they switch roles. They should continue to switch roles as time allows.

Conversing: MLR2 Collect and Display. As students work on this activity, listen for and collect the language students use to explain how they solved the equations. Write the students’ words and phrases on a visual display. As students review the visual display, ask them to clarify the meaning of a phrase. For example, a phrase such as, “Add 4 and then raise to the third power,” can be clarified by restating it as, “Add 4 to each side of the equation. Then raise each side of the equation to the third power.” This will help students use the mathematical language necessary to precisely explain how to solve equations involving cube roots.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer collaboration. When students share the equations they selected with a partner, display sentence frames to support conversation such as: “I chose _____ (equation) because . . .”, “I noticed _____ so I chose . . .”, and “I agree/disagree because . . . .”
Supports accessibility for: Language; Social-emotional skills

### Student Facing

Here are a lot of equations:

• $$\sqrt{t+4} = 3$$
• $$\text-10 = \text-\sqrt{a}$$
• $$\sqrt{3-w} - 4 = 0$$
• $$\sqrt{z} + 9 = 0$$
• $$\sqrt{r^3 - 19} = 2$$
• $$5 - \sqrt{k+1} = \text-1$$
• $$\sqrt{p+4} - 2 = 1$$
• $$6-\sqrt{b} = 0$$
• $$\sqrt{2n} + 3 = \text-5$$
• $$4+\sqrt{\text-m} + 4 = 6$$
• $$\text-\sqrt{c} = 5$$
• $$\sqrt{s-7} + 3 = 0$$
1. Without solving, identify 3 equations that you think would be the least difficult to solve and 3 equations that you think would be the most difficult to solve. Be prepared to explain your reasoning.
2. Choose 4 equations and solve them. At least one should be from your “least difficult” list and at least one should be from your “most difficult” list.

### Student Response

For access, consult one of our IM Certified Partners.

### Student Facing

#### Are you ready for more?

All of these equations were equivalent to equations that could be written in the form $$\sqrt{ax+b}+c=0$$ for some constants $$a$$, $$b$$, and $$c$$. Find a formula that would solve any such equation for $$x$$ in terms of $$a$$, $$b$$, and $$c$$.

### Student Response

For access, consult one of our IM Certified Partners.

### Anticipated Misconceptions

If students have trouble getting started on their more difficult equations, remind them of strategies they used with square roots, such as isolating the root first. Some students may find it helpful to estimate an answer and then check it, in order to get a general idea of what the answer may be and the calculations involved in finding it. If needed, students can look at the graph of $$y=\sqrt{x}$$ from the previous activity, or $$y=x^3$$ from earlier in the lesson.

Students who don’t reach a solution should say where they got stuck and what they tried when they share their work with their partner. They can then try finishing the problem together.

### Activity Synthesis

Much discussion takes place between partners. Invite students to share how they solved the equations they chose. Here are some questions for discussion:

• “How is the process of solving these equations the same as solving equations that involve square roots? How is it different?” (Solving these equations is different because they require cubing rather than squaring. It is also different because cube roots aren’t always positive, so the extra step to check that a positive number isn’t set equal to a negative number isn’t necessary. Solving is the same because it requires isolating the radical expression and then using exponents to make a new equation without radicals.)
• “Describe any difficulties you experienced and how you resolved them.” (I cubed a negative number and forgot that the result would be negative; when I plugged in my answer to check it, I saw that I must have done something wrong, so I went back and fixed it. I wasn’t sure how to get started on one of my difficult equations, but I remembered that I should try to isolate the cube root, and then I could see what to do.)

## Lesson Synthesis

### Lesson Synthesis

In this lesson, students have reasoned about cubes and cube roots to solve equations. Arrange students in groups of 2. Here are some questions for discussion:

• “What are some differences between the graphs of $$y=x^2$$ and $$y=x^3$$, and what do those differences say about solutions to equations like $$x^2=5$$, $$x^2=\text-5$$, $$x^3=5$$, and $$x^3=\text- 5$$?” (The shape of the graph of $$y=x^2$$ means that $$x^2=5$$ has two solutions, $$\sqrt{5}$$ and $$\text- \sqrt{5}$$, and $$x^2=\text-5$$ has no solutions. On the other hand, the shape of the graph of $$y=x^3$$ tells us that $$x^3=5$$ has exactly one solution, $$\sqrt{5}$$, and $$x^3=\text- 5$$ also has one solution, $$\sqrt{\text-5}$$.)
• “Write an equation involving a cube root for your partner to solve and then explain how you solved their equation.” (For example, $$\sqrt{1-x}+8=\text-2$$.)

## 8.5: Cool-down - Cube It (5 minutes)

### Cool-Down

For access, consult one of our IM Certified Partners.

## Student Lesson Summary

### Student Facing

Every number has exactly one cube root. You can see this by looking at the graph of $$y=x^3$$.

If $$y$$ is any number, for example, -4, then we can see that $$y= \text{-}4$$ crosses the graph in one and only one place, so the equation $$x^3 = \text{-}4$$ will have the solution $$\text{-}\sqrt{4}$$. This is true for any number $$a$$: $$y = a$$ will cross the graph in exactly one place, and $$x^3 = a$$ will have one solution, $$\sqrt{a}$$.

In an equation like $$\sqrt{t} + 6 = 0$$, we can isolate the cube root and cube each side:
\begin{align}\sqrt{t} + 6 &= 0\\ \sqrt{t} &= \text{-}6\\ t &= (\text{-}6)^3 \\ t &= \text{-}216 \end{align}