\[ \text { SECTION -C } \] Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. \( \begin{array}{lll}\text { (i) } 243 & \text { (ii) } 72\end{array} \)

To find the smallest number by which each of the given numbers must be multiplied to obtain a perfect cube, we first need to factor each number into its prime factors and then analyze the exponents of these prime factors. ### (i) For the number 243: 1. **Prime Factorization**: \[ 243 = 3^5 \] 2. **Analyzing the Exponents**: The exponent of 3 in the factorization is 5. For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3. The nearest multiple of 3 that is greater than or equal to 5 is 6. Therefore, we need to increase the exponent of 3 from 5 to 6. 3. **Finding the Multiplier**: To achieve this, we need to multiply by \(3^{6-5} = 3^1 = 3\). Thus, the smallest number by which 243 must be multiplied to obtain a perfect cube is **3**. ### (ii) For the number 72: 1. **Prime Factorization**: \[ 72 = 2^3 \times 3^2 \] 2. **Analyzing the Exponents**: - The exponent of 2 is 3, which is already a multiple of 3. - The exponent of 3 is 2. The nearest multiple of 3 that is greater than or equal to 2 is 3. Therefore, we need to increase the exponent of 3 from 2 to 3. 3. **Finding the Multiplier**: To achieve this, we need to multiply by \(3^{3-2} = 3^1 = 3\). Thus, the smallest number by which 72 must be multiplied to obtain a perfect cube is also **3**. ### Summary: - For 243, the multiplier is **3**. - For 72, the multiplier is **3**.