<p>This is a more difficult problem than most.</p>
<p>Let N be the number. We have N = 12<em>k, and N = n</em>n*n.</p>
<p>Factor the number 12 – the “best” factorization is that as the product of primes. You have 12 = 2<em>2</em>3. The factorization of a number into powers of primes is a unique factorization.</p>
<p>For N to be a cube we need at least another factor of 2, and two factors of 3. Then we would have N = (2<em>2</em>2)<em>(3</em>3<em>3). Or N =(2</em>3)^^3 = 6^^3. So k = 2<em>3</em>3</p>
<p>To try out your understanding consider a variant of the problem. N = 14<em>k and N is again a cube. Here 14 = 2</em>7. That’s the unique factorization of 14 into primes. So k would need to be (minimally) (2<em>2)</em>(7*7).</p>
<p>A simpler variant is this. Suppose N=6<em>k, and instead of N being a cube it’s a square. So first we write 6 = 2</em>3. And now k needs to be a product of 2<em>3; this so N=(2</em>2)<em>(3</em>3), And for this example k = 2*3 = 6.</p>
<p>12k = 2 x 2 x 3 x k. you want to have a m^3 from a 12k, so at least you should have a (2^3)x(3^3) from the 12k, since 12 is already a multiple of 2 and 3, right? that is why k is what makes up (2^3)x(3^3) together with 12, which is 2 x 3 x 3 = 18</p>
<p>Now here are some integers that are not perfect cubes:</p>
<p>2<em>2=4 is not a perfect cube, but if we multiply by 2 we get a perfect cube:
2</em>2*2=8</p>
<p>2<em>2</em>2<em>3</em>3=72 is not a perfect cube, but multiplying by 3 makes it a perfect cube:
2<em>2</em>2<em>3</em>3*3=216</p>
<p>Ok. One more example, I’m going to pick a random integer, say 84, and find the least integer I need to multiply by to get a perfect cube. We start by factoring 84:</p>