Start with p/q, and choose a positive rational number x/y so that $\tfrac{p}{q}\cdot(\tfrac{x}{y})^3=\tfrac{P}{Q}$ is strictly between 1 and 2. (This is equivalent to $\sqrt[3]{\tfrac{q}{p}}<\tfrac{x}{y}<\sqrt[3]{\tfrac{2q}{p}}$, and there must exist a rational number in this interval.) Then we can verify that \[ \frac{P}{Q} = \frac{(P+Q)^3+(2P-Q)^3}{(P+Q)^3+(2Q-P)^3} \](where all 4 of those cubes are positive because $1 < \tfrac{P}{Q} < 2$), which means \[ \frac{p}{q} = \frac{y^3(P+Q)^3+y^3(2P-Q)^3}{x^3(P+Q)^3+x^3(2Q-P)^3} \] and we're done.

Note that (m + n)3 + (2m - n)3 = 9m(m2 - mn + n2) and (m + n)3 + (2n - m)3 = 9n(m2 - mn + n2). So a = m + n, b = 2m - n, c = m + n, d = 2n - m gives m/n. We require a, b, c, d positive. Obviously a and c are positive, but b and d are only positive if 2m > n and 2n > m, in other words if m/n belongs to the open interval (1/2, 2). For k > 4, we have (k+1)3/k3 < 2. So consider the sequence 53/43, 63/43, 73/43, ... it obviously tends to infinity and the ratio between successive terms is less than 2. So at least one member belongs to each of the intervals (2, 4), (4, 8), (8, 16), ... . Now any positive rational h < 1/2 must belong to some interval [1/2n+1, 1/2n], but we can find an integer k such that k3/43 belongs to (2n, 2n+1) so h k3/43 belongs to (1/2, 2) and hence can be written as (a3 + b3)/(c3 + d3). So h = ( (4a)3 + (4b)3)/( (kc)3 + (kd)3). Similarly for any rational h > 2. davron ft. kalva

note that for there exists (possibly negative) integer $\alpha$ and $1\leq q \leq \frac{5}{4}<2$ such that $q=\left(\frac{5}{4}\right)^{3\alpha} q'$. It is easy to see that if $q'$ can be represented in the required form, so can $q$. Let $q=\frac{x}{y}$, now we note $q=\frac{(x+y)^3+(2x-y)^3}{(x+y)^3+(2x-y)^3}$. Since $1\leq q \leq 2$, we have $2x>y$ and $2y>x$, ensuring that all the numbers are positive.

It seems the key to this identity is to use that identity with the cubes. I searched for such an identity for 2 hours, but found nothing. Can someone provide motivation or explain how to think of such an identity?

viperstrike wrote: It seems the key to this identity is to use that identity with the cubes. I searched for such an identity for 2 hours, but found nothing. Can someone provide motivation or explain how to think of such an identity? I tried to find something for 1/2, failed. But after a while I found something for 2/3 and it was (5^3+1^3)/(5^3+4^3) and so I conjuectured that the 5^3 was because of that 2+3 = 5. After that it was easy to test things like 3/4, 3/5 and find the correct identity.

Note that \[\frac{x}{y}=\frac{(x+y)^3+(2x-y)^3}{(x+y)^3+(2y-x)^3}.\]This construction works for all $\frac{1}{2}< \frac{x}{y}< 2.$ Now for any rational not in this range we can scale it by $\frac{p}{q}$ to make it in that range, and then do \[\frac{p^3(x+y)^3+p^3(2x-y)^3}{q^3(x+y)^3+q^3(2y-x)^3}=\frac{p^3}{q^3}\cdot \frac{x}{y}.\]

Let the rational number be $\frac{m}{n}$ and WLOG $\frac{7}{13}<\frac{m}{n}<\frac{13}{7}$ by scaling (we can multiply/divide by any cube of an integer to obtain all rationals). Then take \[\frac{m}{n}=\frac{(13m-7n)^3+(11m+7n)^3}{(13n-7m)^3+(11n+7m)^3}\]which finishes. $\blacksquare$

spent so long trying to get contradiction to work this problem gave me hell, i think the construction is pretty motivated once you think of sum of cubes though Note that any fraction $\frac xy$ with $x, y \in \mathbb{Z}$ and $\frac12 < \frac xy < 2$ can be represented as \[\frac{(x+y)^3+(2x-y)^3}{(x+y)^3+(2y-x)^3};\]noting that we may scale appropriately by any cube of a number in $\mathbb{Q}$, we are done. $\square$

Note that \[ \frac{p}{q} = \frac{a^3+b^3}{c^3+d^3} = \frac{(a+b)(a^2 - ab + b^2)}{(c+d)(c^2 - cd + d^2)} \]So to make our life easier, let us set $\frac{a+b}{c+d} = \frac{p}{q}$ and see if we can find $a$ and $b$ to make $a^2 - ab + b^2 = c^2 - cd + d^2$. Note that we can set $a+b = kp$ and $c+d = kq$. Then trying to use $a+b$, we must have \[ (a+b)^2 - 3a(a+b) + 3a^2 = (a+b)^2 - 3ab = a^2 - ab + b^2 = c^2 - cd + d^2 = (c+d)^2 - 3c(c+d) + 3c^2 \]Thus, we must have that \[ (kp)^2 - (kq)^2 = 3a(kp) - 3c(kq) + 3c^2 - 3a^2 \]In particular, then $3 \mid k$. So we could try $k = 3$. Then we end up wih \[ 9p^2 - 3a \cdot 3p + 3a^2 = 9q^2 - 3c \cdot 3q + 3c^2 \implies 3p^2 - 3ap + a^2 = 3q^2 - 3cq + c^2 \]Now, note we should probably choose $a$ in the form of $mp + nq$, and in particular then we just want the LHS to be symmetric in $p$ and $q$. Let $m = 1$ to cancel out the $3p^2$, and then we find that $n = 1$ for it to be symmetric. This means that a solution is $a = p+q$, $b = 2p - q$, $c = p+q$, $d = 2q - p$ and we can easily check that this works. But this only works if $\frac{1}{2} < \frac{p}{q} < 2$. However, we can always scale $\frac{p}{q}$ down by some cube $\frac{m^3}{n^3}$ to be in this range, and them scale back up to get our solution. $\blacksquare$

Set $b + d = a = c$. Then we know $b^2 -ab +a^2 - d^2 +ad - a^2= (b + d - a)(b - d) = 0$, so we can write $\frac{a^3 + b^3}{a^3 + d^3} = \frac{a + b}{a + d} = \frac{a+ b}{2a - b} = \frac{\frac 32a + x}{\frac 32a - x}$ for some $x$. Thus if we desire to represent $\frac pq$, we can write $\frac pq = \frac{6p}{6q}$, so set $a = 2p + 2q, x = 3p - 3q$, which yields the desired results. This gives $b = 4p - 2q, q = 4q - 2p$. This only then works if $\frac 12 < \frac pq < 2$, but then this is easily fixed by scaling by cubes.