7 8) Written as a product of its prime factors, \( 588=2^{2} \times 3 \times 7^{2} \). (a) Find the smallest positive integer value of \( k \) such that \( 588 k \) is a perfect cube. Answer \( k= \) \( \qquad \) (b) Express 90 as a product of its prime factors.

To solve the problem, we will break it down into two parts as requested. ### Part (a) We need to find the smallest positive integer value of \( k \) such that \( 588k \) is a perfect cube. 1. **Prime Factorization of 588**: The prime factorization of \( 588 \) is given as: \[ 588 = 2^{2} \times 3^{1} \times 7^{2} \] 2. **Perfect Cube Condition**: For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3. 3. **Current Exponents**: - For \( 2^{2} \): The exponent is \( 2 \). - For \( 3^{1} \): The exponent is \( 1 \). - For \( 7^{2} \): The exponent is \( 2 \). 4. **Adjusting Exponents**: - To make \( 2^{2} \) a perfect cube, we need \( 2^{2} \) to become \( 2^{3} \). We need \( 1 \) more factor of \( 2 \). - To make \( 3^{1} \) a perfect cube, we need \( 3^{1} \) to become \( 3^{3} \). We need \( 2 \) more factors of \( 3 \). - To make \( 7^{2} \) a perfect cube, we need \( 7^{2} \) to become \( 7^{3} \). We need \( 1 \) more factor of \( 7 \). 5. **Calculating \( k \)**: Therefore, we need: \[ k = 2^{1} \times 3^{2} \times 7^{1} \] Now, we calculate \( k \): \[ k = 2^{1} \times 3^{2} \times 7^{1} = 2 \times 9 \times 7 \] First, calculate \( 2 \times 9 = 18 \). Then, calculate \( 18 \times 7 = 126 \). Thus, the smallest positive integer value of \( k \) such that \( 588k \) is a perfect cube is: \[ \boxed{126} \] ### Part (b) Now, we need to express \( 90 \) as a product of its prime factors. 1. **Finding Prime Factors of 90**: - Start by dividing \( 90 \) by \( 2 \) (the smallest prime): \[ 90 \div 2 = 45 \] - Next, divide \( 45 \) by \( 3 \): \[ 45 \div 3 = 15 \] - Finally, divide \( 15 \) by \( 3 \) again: \[ 15 \div 3 = 5 \] - Now, \( 5 \) is a prime number. 2. **Writing the Prime Factorization**: Therefore, the prime factorization of \( 90 \) is: \[ 90 = 2^{1} \times 3^{2} \times 5^{1} \] Thus, \( 90 \) expressed as a product of its prime factors is: \[ \boxed{2^{1} \times 3^{2} \times 5^{1}} \]