The number of pairs of positive integers \(x\), \(y\) which solve the equation \[x^3 + 6x^2y + 12xy^2 + 8y^3 = 2^{30}\] is

- \(0\),
- \(2^6\),
- \(2^9-1\),
- \(2^{10}+2\).

We will take it the question means ordered pairs, so that \((x,y)\) is counted as different from \((y,x)\) if \(x \neq y\).

First we factorise the equation to get: \[(x+2y)^3=2^{30},\] which when we cube root both sides gives us \[x+2y=2^{10}.\]

The function \(y=x^3\) is always increasing, so we don’t lose any solutions by doing this.

Let’s try to systematically list the pairs of positive integers (\(x\),\(y\)) that would satisfy this equation.

The value \(x=1\) does not work, because \(\dfrac{2^{10}-1}{2}\) is not an integer. Clearly we require \(x\) to be even for a solution.

So we have \[(2,2^9-1), (4,2^9-2), (6,2^9-3),\dots, (2^{10}-4,2),(2^{10}-2,1).\]

So \(y\) takes every value from \(1\) to \(2^9-1\), which gives us that there are \(2^9-1\) pairs of positive integers that satisfy the equation, and the answer is (c).