To show that $3375$ is a perfect cube and find the number whose cube it is, we proceed as follows:

Step 1: Prime Factorization of $3375$

We factorize $3375$ into its prime factors: $3375 \div 3 = 1125 \quad (\text{since } 3375 \text{ is divisible by } 3)$ $1125 \div 3 = 375$ $375 \div 3 = 125$ $125 \div 5 = 25$ $25 \div 5 = 5$ $5 \div 5 = 1$

Thus, the prime factorization of $3375$ is: $3375 = 3^3 \cdot 5^3$

Step 2: Grouping into Triplets

A number is a perfect cube if all its prime factors can be grouped into triplets. Here: $3^3 \cdot 5^3$ Both $3$ and $5$ appear in powers of $3$, so $3375$ is a perfect cube.

Step 3: Cube Root of $3375$

The cube root of $3375$ is given by: [ \sqrt[3]{3375} = \sqrt[3]{3^3 \cdot 5^3} = 3 \cdot 5 = 15 ]

Final Answer

• $3375$ is a perfect cube.
• The number whose cube is $3375$ is $15$.

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Do you have any questions or would you like further details?
Here are some related questions you might find useful:

1. What is the general method for identifying perfect cubes?
2. How do you calculate cube roots of larger numbers manually?
3. Can you explain why grouping into triplets determines perfect cubes?
4. How can we verify a perfect cube using a calculator?
5. What is the significance of prime factorization in cube root calculations?

Tip: Always check divisibility by smaller primes (2, 3, 5, etc.) when performing prime factorization.