Math worksheet focusing on cubes and cube roots, with questions on perfect cubes, cube values, and properties of numbers.

Let's solve each question one by one with explanations.

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1. Which is the smallest three-digit perfect cube?

We need to find the smallest perfect cube that is a 3-digit number (i.e., between 100 and 999).

Let’s check cubes of small integers:

- $ 4^3 = 64 $ → too small (2-digit)
- $ 5^3 = 125 $ → 3-digit ✔
- $ 6^3 = 216 $
- $ 7^3 = 343 $
- $ 8^3 = 512 $
- $ 9^3 = 729 $
- $ 10^3 = 1000 $ → 4-digit

So, the smallest 3-digit perfect cube is 125.

✔ Answer: (a) 125

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2. Which is the greatest three-digit perfect cube?

From above:
- $ 9^3 = 729 $
- $ 10^3 = 1000 $ → too big

So, the largest 3-digit perfect cube is 729.

✔ Answer: (c) 729

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3. Which of the following is not a perfect cube?

Check each option:

- (a) 1 → $ 1^3 = 1 $ → perfect cube ✔
- (b) 9 → Is there an integer $ x $ such that $ x^3 = 9 $?
$ 2^3 = 8 $, $ 3^3 = 27 $ → no → not a perfect cube ✘
- (c) 8 → $ 2^3 = 8 $ → perfect cube ✔
- (d) 27 → $ 3^3 = 27 $ → perfect cube ✔

✔ Answer: (b) 9

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4. The cube of 4 is _________.

$ 4^3 = 4 \times 4 \times 4 = 64 $

✔ Answer: (d) 64

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5. The value of $ 5^3 $ is _________.

$ 5^3 = 5 \times 5 \times 5 = 125 $

✔ Answer: (a) 125

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6. The cube of an even number is always _________.

Even numbers are divisible by 2.

Let’s take an example: $ 2^3 = 8 $, $ 4^3 = 64 $, $ 6^3 = 216 $ — all even.

Why? Because if $ n $ is even, then $ n = 2k $, so $ n^3 = (2k)^3 = 8k^3 $, which is divisible by 2 → even.

So cube of even number is always even.

✔ Answer: (b) even number

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7. The cube of an odd number is always _________.

Odd numbers: $ 1, 3, 5, \dots $

$ 1^3 = 1 $, $ 3^3 = 27 $, $ 5^3 = 125 $ — all odd.

If $ n $ is odd, $ n = 2k+1 $, then $ n^3 = (2k+1)^3 $ — expansion gives odd result.

So cube of odd number is odd.

✔ Answer: (a) odd number

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8. Each prime factor appears ________ times in its cube?

Suppose a number has prime factorization:
$ n = p_1^{a} \cdot p_2^{b} \cdot \ldots $

Then $ n^3 = p_1^{3a} \cdot p_2^{3b} \cdot \ldots $

So each prime factor appears 3 times as many times as it did in the original number.

For example: $ 6 = 2^1 \cdot 3^1 $, $ 6^3 = 2^3 \cdot 3^3 $

So each prime factor appears 3 times in the exponent.

✔ Answer: (b) 3

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9. Which of the following is Hardy-Ramanujan Number?

The Hardy-Ramanujan number is 1729, famous for being the smallest number expressible as the sum of two cubes in two different ways:

$$
1729 = 1^3 + 12^3 = 9^3 + 10^3
$$

So among the options:

- (a) 1724
- (b) 1725
- (c) 1727
- (d) 1729 → ✔

✔ Answer: (d) 1729

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10. By which smallest natural number must 392 be multiplied so that the product is a perfect cube?

We need to find the smallest number $ x $ such that $ 392 \times x $ is a perfect cube.

First, factorize 392.

$$
392 \div 2 = 196 \\
196 \div 2 = 98 \\
98 \div 2 = 49 \\
49 = 7 \times 7
$$

So:

$$
392 = 2^3 \times 7^2
$$

For a number to be a perfect cube, all exponents in prime factorization must be multiples of 3.

Currently:
- $ 2^3 $ → already good (exponent 3)
- $ 7^2 $ → needs one more 7 to become $ 7^3 $

So we need to multiply by 7 to make $ 7^3 $

Then:
$$
392 \times 7 = 2^3 \times 7^3 = (2 \times 7)^3 = 14^3
$$

✔ Answer: (c) 7

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✔ Final Answers:

1. (a) 125
2. (c) 729
3. (b) 9
4. (d) 64
5. (a) 125
6. (b) even number
7. (a) odd number
8. (b) 3
9. (d) 1729
10. (c) 7

Let me know if you'd like a printable version or explanation in simpler terms!