Check whether 1728 is a perfect cube by using prime factorisation.

Answer

To check whether 1728 is a perfect cube using prime factorization, we can follow these steps:

### Step 1: Prime Factorization of 1728
We start by finding the prime factors of 1728. We can do this by dividing the number by the smallest prime number until we reach 1.

1. Divide 1728 by 2 (the smallest prime number):
\[ 1728 \div 2 = 864 \]
2. Divide 864 by 2:
\[ 864 \div 2 = 432 \]
3. Divide 432 by 2:
\[ 432 \div 2 = 216 \]
4. Divide 216 by 2:
\[ 216 \div 2 = 108 \]
5. Divide 108 by 2:
\[ 108 \div 2 = 54 \]
6. Divide 54 by 2:
\[ 54 \div 2 = 27 \]
7. Now, 27 is not divisible by 2, so we move to the next prime number, which is 3:
\[ 27 \div 3 = 9 \]
8. Divide 9 by 3:
\[ 9 \div 3 = 3 \]
9. Finally, divide 3 by 3:
\[ 3 \div 3 = 1 \]

So, the prime factorization of 1728 is:
\[ 1728 = 2^6 \times 3^3 \]

### Step 2: Checking for Perfect Cube
A number is a perfect cube if all the exponents in its prime factorization are multiples of 3.

- The exponent of 2 is 6, which is a multiple of 3 (since \(6 \div 3 = 2\)).
- The exponent of 3 is 3, which is also a multiple of 3 (since \(3 \div 3 = 1\)).

Since both exponents are multiples of 3, we can conclude that 1728 is a perfect cube.

### Step 3: Finding the Cube Root
To find the cube root of 1728, we take the cube root of each prime factor raised to its respective exponent:
\[ \sqrt[3]{1728} = \sqrt[3]{2^6} \times \sqrt[3]{3^3} = 2^{6/3} \times 3^{3/3} = 2^2 \times 3^1 = 4 \times 3 = 12 \]

Thus, the cube root of 1728 is 12.

### Conclusion
1728 is a perfect cube, and its cube root is 12.

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