The product of multiple perfect cubes is also a perfect cube. Proof for two: n^3 * p^3 = nnnppp = npnpnp = (np)^3 And any integer whose exponent is a multiple of 3 is a perfect cube. We will use this here:
Prime factorize 3240: 3240 = 405 * 2^3 = 2^3 * 3^4 * 5^1 We need to multiply this by k, to make all the exponents divisible by 3 The exponents not divisible by 3 are 4 and 1. So let's fix that: 2^3 * 3^4 * 3*2 * 5^1 * 5^2 So, k is 3^2 * 5^2 = 225 (3240*225)^(1/3) = 90
