Square Root and Cube Root Practice worksheet with 20 problems and a colorful bird illustration.

Let's solve each problem on the "Square Root and Cube Root Practice" worksheet step by step.

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Instructions:

Find each square root or cube root.

We'll go through each problem:

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1. √81
$ 9 \times 9 = 81 $ → $ \sqrt{81} = 9 $

2. √64
$ 8 \times 8 = 64 $ → $ \sqrt{64} = 8 $

3. √16
$ 4 \times 4 = 16 $ → $ \sqrt{16} = 4 $

4. ∛216
$ 6 \times 6 \times 6 = 216 $ → $ \sqrt[3]{216} = 6 $

5. √49
$ 7 \times 7 = 49 $ → $ \sqrt{49} = 7 $

6. √225
$ 15 \times 15 = 225 $ → $ \sqrt{225} = 15 $

7. √27
$ 27 $ is not a perfect square.
But $ \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} $
However, since this seems to be a basic practice sheet, it might expect an approximate value or recognize it’s not a whole number.
But check: Is there a typo? Wait — maybe it's meant to be a cube root? No, it's written as √27 (square root).
So: $ \sqrt{27} \approx 5.196 $, but likely not expected here.
Let’s double-check:
Actually, 27 is not a perfect square, so unless simplified, it's irrational.
But perhaps it's a mistake? Wait — look at #8: √100 = 10 — that's fine.

So for now: $ \sqrt{27} $ is not a whole number, but we can simplify:
$$
\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}
$$
But if only whole numbers are expected, maybe this is a typo?

Alternatively, could it be cube root of 27? That would be $ \sqrt[3]{27} = 3 $.
But the symbol is √, not ∛. So probably just leave it as $ \sqrt{27} $, but let’s assume the worksheet expects exact values. Since 27 is not a perfect square, maybe it's a trick?

Wait — maybe I misread. Let's check all carefully.

Actually, looking again:
Problem 7: √27 → no perfect square, so likely not intended unless simplification is allowed.

But let's continue and see pattern.

Maybe it's a typo? Or perhaps it's supposed to be ∛27? But no, it's written as √27.

So answer: $ \sqrt{27} = 3\sqrt{3} $, but if only integers are expected, then perhaps skip or note it's not a whole number.

But wait — look at #13: ∛343 → that's a cube root.

So perhaps #7 is indeed √27, and it's acceptable to write $ 3\sqrt{3} $, but since this is a practice sheet with mostly whole numbers, maybe it's a mistake?

Let's keep going.

8. √100
$ 10 \times 10 = 100 $ → $ \sqrt{100} = 10 $

9. √169
$ 13 \times 13 = 169 $ → $ \sqrt{169} = 13 $

10. √256
$ 16 \times 16 = 256 $ → $ \sqrt{256} = 16 $

11. ∛1,000
$ 10 \times 10 \times 10 = 1000 $ → $ \sqrt[3]{1000} = 10 $

12. ∛512
$ 8 \times 8 \times 8 = 512 $ → $ \sqrt[3]{512} = 8 $

13. ∛343
$ 7 \times 7 \times 7 = 343 $ → $ \sqrt[3]{343} = 7 $

14. √324
$ 18 \times 18 = 324 $ → $ \sqrt{324} = 18 $

15. √125
$ 125 $ is not a perfect square.
$ \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} $ → not a whole number.

But again, maybe it's a typo? Could it be ∛125?
$ \sqrt[3]{125} = 5 $, since $ 5^3 = 125 $

But the symbol is √, not ∛ → so it's square root.

So $ \sqrt{125} = 5\sqrt{5} \approx 11.18 $

But again, if only whole numbers, maybe it's meant to be cube root?

Let’s look at #16: √1,331

16. √1,331
$ 1,331 $ is not a perfect square.
$ \sqrt{1,331} \approx 36.5 $, but check: $ 36^2 = 1296 $, $ 37^2 = 1369 $ → not perfect square.

But wait — $ 11^3 = 1331 $ → so $ \sqrt[3]{1331} = 11 $

So likely, #16 should be ∛1,331, not √1,331.

This suggests a pattern: some problems may have incorrect symbols.

Let's re-express the list with possible corrections.

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Wait — let's look at the original image description:
It says “Square Root and Cube Root Practice”, so both types are included.

But looking at the layout:

- Problems 1–6: Square roots
- 7–8: Square roots
- 9–10: Square roots
- 11–12: Cube roots
- 13–16: Mixed?
- 17–20: Likely cube roots?

Wait — let's count:

| # | Expression | Type |
|---|------------------|-------------|
| 1 | √81 | √ |
| 2 | √64 | √ |
| 3 | √16 | √ |
| 4 | ∛216 | ∛ |
| 5 | √49 | √ |
| 6 | √225 | √ |
| 7 | √27 | √ |
| 8 | √100 | √ |
| 9 | √169 | √ |
|10 | √256 | √ |
|11 | ∛1,000 | ∛ |
|12 | ∛512 | ∛ |
|13 | ∛343 | ∛ |
|14 | √324 | √ |
|15 | √125 | √ |
|16 | √1,331 | √ |
|17 | √729 | √ |
|18 | ∛1,728 | ∛ |
|19 | √196 | √ |
|20 | √361 | √ |

Now analyze:

- Perfect squares:
- 81 → 9² → 9
- 64 → 8² → 8
- 16 → 4² → 4
- 49 → 7² → 7
- 225 → 15² → 15
- 100 → 10² → 10
- 169 → 13² → 13
- 256 → 16² → 16
- 324 → 18² → 18
- 196 → 14² → 14
- 361 → 19² → 19
- 729 → 27² → 27

- Perfect cubes:
- 216 → 6³ → 6
- 1000 → 10³ → 10
- 512 → 8³ → 8
- 343 → 7³ → 7
- 1728 → 12³ → 12

So now we see:

- #7: √27 → Not a perfect square → $ \sqrt{27} = 3\sqrt{3} $
- #15: √125 → $ \sqrt{125} = 5\sqrt{5} $
- #16: √1,331 → $ \sqrt{1,331} $ is not a perfect square → but $ 1,331 = 11^3 $ → so $ \sqrt[3]{1,331} = 11 $

Similarly:
- #17: √729 → $ \sqrt{729} = 27 $ → because $ 27^2 = 729 $

Wait! $ 27^2 = 729 $? Let's check:
$ 27 \times 27 = (20+7)^2 = 400 + 2×20×7 + 49 = 400 + 280 + 49 = 729 $ → Yes!

So √729 = 27

But earlier I thought √125 was problematic — but it's still not a perfect square.

However, notice:
- #18: ∛1,728 → $ 12^3 = 1,728 $ → so $ \sqrt[3]{1,728} = 12 $

So likely, some of the square roots are not perfect squares, but others are.

Let’s go through all 20 and give correct answers.

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✔ Final Answers:

1. $ \sqrt{81} = 9 $
2. $ \sqrt{64} = 8 $
3. $ \sqrt{16} = 4 $
4. $ \sqrt[3]{216} = 6 $
5. $ \sqrt{49} = 7 $
6. $ \sqrt{225} = 15 $
7. $ \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} $ → Not a whole number
But if only integers expected, maybe it's a typo?
However, $ \sqrt{27} \approx 5.196 $
But perhaps leave as $ 3\sqrt{3} $

8. $ \sqrt{100} = 10 $
9. $ \sqrt{169} = 13 $
10. $ \sqrt{256} = 16 $
11. $ \sqrt[3]{1000} = 10 $
12. $ \sqrt[3]{512} = 8 $
13. $ \sqrt[3]{343} = 7 $
14. $ \sqrt{324} = 18 $
15. $ \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} $ → not whole
16. $ \sqrt{1331} $ → not a perfect square
$ 36^2 = 1296 $, $ 37^2 = 1369 $ → so between them → not integer
But $ 1331 = 11^3 $ → so $ \sqrt[3]{1331} = 11 $ → likely typo in symbol
So if it were ∛1331, answer is 11
But as √1331, it's not rational

17. $ \sqrt{729} = 27 $ → because $ 27^2 = 729 $
18. $ \sqrt[3]{1728} = 12 $ → because $ 12^3 = 1728 $
19. $ \sqrt{196} = 14 $ → $ 14^2 = 196 $
20. $ \sqrt{361} = 19 $ → $ 19^2 = 361 $

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🔍 Observations:

- Problems 7, 15, 16 are not perfect squares.
- But #16: √1331 is likely meant to be ∛1331 = 11, since 1331 is a perfect cube.
- Similarly, #15: √125 → if it were ∛125 = 5, that would make sense.
- #7: √27 → if it were ∛27 = 3, that would fit.

So it's possible there are symbol errors in the worksheet.

Let’s assume the worksheet has typographical errors in some cases.

But based on what’s written:

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✔ Corrected Answers Based on Given Symbols:

1. $ \sqrt{81} = 9 $
2. $ \sqrt{64} = 8 $
3. $ \sqrt{16} = 4 $
4. $ \sqrt[3]{216} = 6 $
5. $ \sqrt{49} = 7 $
6. $ \sqrt{225} = 15 $
7. $ \sqrt{27} = 3\sqrt{3} $ ≈ 5.196
8. $ \sqrt{100} = 10 $
9. $ \sqrt{169} = 13 $
10. $ \sqrt{256} = 16 $
11. $ \sqrt[3]{1000} = 10 $
12. $ \sqrt[3]{512} = 8 $
13. $ \sqrt[3]{343} = 7 $
14. $ \sqrt{324} = 18 $
15. $ \sqrt{125} = 5\sqrt{5} $ ≈ 11.18
16. $ \sqrt{1331} \approx 36.5 $ (not exact)
17. $ \sqrt{729} = 27 $
18. $ \sqrt[3]{1728} = 12 $
19. $ \sqrt{196} = 14 $
20. $ \sqrt{361} = 19 $

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✔ If We Assume Typo Corrections (Most Likely):

Many of these numbers are perfect cubes, so likely the symbols were mixed up.

Let’s suppose:

- #7: ∛27 = 3 (since 3³=27)
- #15: ∛125 = 5 (5³=125)
- #16: ∛1331 = 11 (11³=1331)

Then the answers become all nice integers.

Also, #17: √729 = 27 is correct as is.

So revised interpretation:

| # | Expression | Correct Answer |
|---|----------------|----------------|
| 1 | √81 | 9 |
| 2 | √64 | 8 |
| 3 | √16 | 4 |
| 4 | ∛216 | 6 |
| 5 | √49 | 7 |
| 6 | √225 | 15 |
| 7 | ∛27 | 3 |
| 8 | √100 | 10 |
| 9 | √169 | 13 |
|10 | √256 | 16 |
|11 | ∛1000 | 10 |
|12 | ∛512 | 8 |
|13 | ∛343 | 7 |
|14 | √324 | 18 |
|15 | ∛125 | 5 |
|16 | ∛1331 | 11 |
|17 | √729 | 27 |
|18 | ∛1728 | 12 |
|19 | √196 | 14 |
|20 | √361 | 19 |

This makes all answers integers, which fits a basic practice worksheet.

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✔ Final Answer (Assuming Symbol Errors Are Fixed):

1. 9
2. 8
3. 4
4. 6
5. 7
6. 15
7. 3
8. 10
9. 13
10. 16
11. 10
12. 8
13. 7
14. 18
15. 5
16. 11
17. 27
18. 12
19. 14
20. 19

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📌 Explanation:

- For perfect squares, find the number that when multiplied by itself gives the radicand.
- For perfect cubes, find the number that when multiplied three times gives the radicand.
- Some entries like √27, √125, √1331 are not perfect squares, so they either need simplification or indicate a typo in the problem.
- Since all other numbers are perfect powers, it's highly likely that some square root symbols should be cube roots.

Thus, the most reasonable solution assumes corrected symbols for problems 7, 15, and 16.

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✔ Final Answer Key (with assumed corrections):

| Problem | Answer |
|--------|--------|
| 1 | 9 |
| 2 | 8 |
| 3 | 4 |
| 4 | 6 |
| 5 | 7 |
| 6 | 15 |
| 7 | 3 |
| 8 | 10 |
| 9 | 13 |
| 10 | 16 |
| 11 | 10 |
| 12 | 8 |
| 13 | 7 |
| 14 | 18 |
| 15 | 5 |
| 16 | 11 |
| 17 | 27 |
| 18 | 12 |
| 19 | 14 |
| 20 | 19 |

Let me know if you'd like the explanation for any specific one!