Now \(90a=b^3.......2*3^2*5*a=b^3\)
So a has to be a multiple of 2^2*3*5^2...
Thus a divided by a factor of 2^2*3*5^2 will be an integer..

I. \(\frac{a}{2^2*3*5}\)
2^2*3*5 is a factor of 2^2*3*5^2....so YES

II. \(\frac{a}{2*3^2*5}\)
2*3^2*5 is NOT a factor of 2^2*3*5^2....so Not necessary

III. \(\frac{a}{2^2*3 *5^2}\)
2^2*3*5^2 is a factor of 2^2*3*5^2....so YES

I and III

Bunuel, no choice matches the answer, so there must be a typo in choice I
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\(90a = b^3\). So, 90a is a perfect cube.

90 = 3*3*5*2*a.

\(a = 3*5^2*2^2\) = 300

scan each option:

I. \(\frac{a}{2^3*3*5}\) = 300 / 120 = no integer.

II. \(\frac{a}{2*3^2*5}\) = 300 / 90 = no integer.

III. \(\frac{a}{2^2*3 *5^2}\) = 300/ 300 = integer .

C is the best answer.

\(90a = b^3\)

90a has to be a perfect cube. How can we make a perfect cube?

If we break down 90 what we get:

90 = 3 * 3*5 * 2 .

how many 3 u have and what's about 5 and 2. u have 3^2 and a 5 and a 2. Still we don't know about a.

when u say something is a perfect cube it means all the prime factors of that number must have to be expressed through cube (3)

\(3^2*5*2\)is not perfect cube. So rest of the elements come from a. remember that 90a is perfect cube. but 90 alone is not a perfect cube.

\(3^2\)..........need another 3

2 .....need \(2^2\)

5.......need \(5^2\)

Therefore this additional properties is the part of a.

\(a = 3 * 2^2 * 5^2 = 300\)

now see: \(90a = 90*300 = b^3\)

We see that 90a must be a perfect cube. Thus 90a must have prime factors whose exponents are multiples of 3. Let’s break 90 into prime factors:

90 = 9 x 10 = 2^1 x 3^2 x 5^1

So in order for 90a to be a perfect cube, the number “a” must contain the product 2^2 x 3^1 x 5^2.

Thus, only III must be an integer.

Answer: C
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