Cubic Diophantine

Given that $\displaystyle x^{3} + y^{3} + \left(\frac{1}{3}\right)^{3} = 3\left(\frac{1}{3}\right)xy$

From $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Given that $LHS = 0$ $(a = x, b = y, c = \frac{1}{3})$ .

Therefore, $a+b+c = 0 \text{ or } a^{2}+b^{2}+c^{2}-ab-bc-ca = 0$

$a+b+c = 0$ doesn't give positive $x \text{ and } y$ .

$a^{2}+b^{2}+c^{2}-ab-bc-ca = 0$

$\displaystyle \frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2} = 0$

We get $a = b \text{ or } b = c \text{ or } c = a$

Which gives $\boxed{\displaystyle x = y = \frac{1}{3}}$ ~~~

Since $x$ and $y$ are positive $x^3$ and $y^3$ must be positive so we can apply AM-GM to get

$x^3 + y^3 + \frac{1}{27} >= xy$

But its given that $x^3 + y^3 + \frac{1}{27} = xy$

We know that AM = GM when all terms on the LHS are equal.

$x^3 = y^3 = \frac{1}{27} \Rightarrow x = y = \frac{1}{3}$

$\Rightarrow x + y = \frac{2}{3} \Rightarrow m + n =5$