Roots and Exponents Simplification

Express in terms of prime factors: $7^a \times (5 \times 7)^{b+1} \times (2^2 \times 5)^{c+2} = 2^{2c+4} \cdot 5^{b+c+3} \cdot 7^{a+b+1}$. For the cube root to be a whole number, powers of 2, 5, 7 must be multiples of 3:

• $2c+4$ divisible by 3
• $b+c+3$ divisible by 3
• $a+b+1$ divisible by 3

Testing option E ($a=3, b=2, c=1$):

• $2(1)+4 = 6$ (divisible by 3)
• $2+1+3 = 6$ (divisible by 3)
• $3+2+1 = 6$ (divisible by 3) All are multiples of 3, so option E is consistent.