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.