Let $x=n^{1/6}$ (i.e. $n=x^6$):
$$2x^3-\frac 3 2x^2+x=y$$
Subtract $y$ and divide both sides by $2$:
$$x^3-\frac 3 4x^2+\frac 1 2x-\frac 1 2y=0$$
In order to reduce this to a depressed cubic, let $x=z+\frac 1 4$:
$$z^3+\frac{5z}{16}+\frac{3}{32}-\frac 1 2y=0$$
In order to turn this into a quadratic, let $z=w+\frac{5}{48w}$:
$$w^3+\frac{3}{32}-\frac 1 2y-\frac{125}{110592w^3}=0$$
Multiply by $w^3$:
$$w^6+\left(\frac{3}{32}-\frac 1 2y\right)w^3-\frac{125}{110592}=0$$
Time to apply the quadratic formula:
$$w^3=\frac{\frac 1 2 y-\frac{3}{32}\pm\sqrt{\left(\frac{3}{32}-\frac 1 2 y\right)^2-\frac{125}{27649}}}{\frac{125}{55296}}$$
Now, there are three complex solutions to this equation. To represent this, I will use $\omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $\omega=1$, one for $\omega=\frac{-1+\sqrt{-3}}{2}$, and one for $\omega=\frac{-1-\sqrt{-3}}{2}$:
$$w=\omega\sqrt[3]{\frac{\frac 1 2 y-\frac{3}{32}\pm\sqrt{\left(\frac{3}{32}-\frac 1 2 y\right)^2-\frac{125}{27649}}}{\frac{125}{55296}}}$$
From here, I leave the rest to you: Solve for $z$ using $z=w-\frac{5}{48w}$, solve for $x$ using $x=z+\frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!
$$T = \left(\frac{1}{AP} - \frac{1}{P}\right)\,Q=\frac{Q}{P}\left(\frac{1}{A}-1\right)\,.$$
Then
$$P=\frac{Q}{T}\left(\frac{1}{A}-1\right)\,.$$
Best Answer
Hint:
the solutions of the system are, for generic $z$: $$ X=\frac{x(1+z)+y(1-z)}{2} \qquad Y=\frac{x(1-z)+y(1+z)}{2} $$ so we can have $X=f(z,x)$ and $Y=f(z,y)$ only for $z=1$.