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To solve this problem, we need to determine the smallest natural number by which \( 72 \) must be multiplied so that the product becomes a perfect cube.

### Step 1: Prime factorization of 72
We begin by breaking 72 into its prime factors:
\[
72 = 2^3 \times 3^2
\]
Here, we see that \( 72 \) is composed of the prime factors 2 and 3, with the exponents 3 and 2, respectively.

### Step 2: Perfect cube condition
For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3.

- The factor \( 2^3 \) is already a perfect cube, as the exponent is 3, which is divisible by 3.
- The factor \( 3^2 \) is not a perfect cube, as the exponent 2 is not divisible by 3. To make it a cube, we need to multiply it by one more factor of \( 3 \) to raise the exponent to \( 3 \).

### Step 3: Solution
To make \( 72 \) a perfect cube, we need to multiply it by \( 3 \), which will make the product:
\[
72 \times 3 = 2^3 \times 3^3 = 216
\]
This is a perfect cube because both exponents are now multiples of 3.

Thus, the smallest natural number by which 72 must be multiplied is \( \boxed{3} \).

Do you want any further details or have any questions?

### Related Questions:
1. What is the prime factorization of 108?
2. How do you identify if a number is a perfect square?
3. How can you determine the smallest number to multiply a given number to make it a perfect square?
4. What are the perfect cubes between 1 and 500?
5. How can you check if a large number is a perfect cube without factorizing?

### Tip:
For any number to be a perfect cube, the exponent of each prime factor in its prime factorization must be divisible by 3.

Find the smallest natural number by which 72 must be multiplied so that the product is a perfect cube.

Learn how to determine the smallest natural number by which 72 must be multiplied so that the product is a perfect cube. This problem involves number theory concepts like prime factorization and the properties of perfect cubes. Suitable for students in Grades 7-9.

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