Find the answer

$\begin{cases} x^4y^5+y^4x^5 = 810 \quad \Rightarrow \begin{cases} (xy)^4(y+x) = 810 \quad \quad \quad \Rightarrow \color{#3D99F6} {x+y = \dfrac{810}{(xy)^4}} \\ (xy)^3(xy^2+yx^2) = 810 \quad \Rightarrow \color{#D61F06}{xy^2+x^2y = \dfrac{810}{(xy)^3}} \end{cases} \\ x^3y^6+ y^3x^6 = 945 \quad \Rightarrow (xy)^3(y^3+x^3) = 945 \quad \quad \quad \Rightarrow \color{#20A900}{x^3 + y^3 = \dfrac{945}{(xy)^3}} \end{cases}$

Now, we have:

$\begin{aligned} (\color{#3D99F6}{x+y})^3 & = \color{#20A900}{x^3+y^3} + 3(\color{#D61F06}{x^2y+xy^2}) \\ \color{#3D99F6}{\frac{810^3}{(xy)^{12}}} & = \color{#20A900}{\frac{945}{(xy)^3}} + 3 \left(\color{#D61F06}{\frac{810}{(xy)^3}}\right) \\ \color{#3D99F6}{\frac{810^3}{(xy)^{9}}} & = \color{#20A900}{945} + \color{#D61F06}{2430} \\ \Rightarrow (xy)^9 & = \frac{810^3}{3375} \\ \color{#69047E}{(xy)^3} & = \color{#69047E}{54} \end{aligned}$

Now, we evaluate:

$\begin{aligned} 2\color{#20A900}{x^3} + \color{#69047E}{(xy)^3} + 2\color{#20A900}{y^3} & = 2\left(\color{#20A900}{\frac{945}{54}} \right) + \color{#69047E}{54} \\ & = \color{#20A900}{35} + \color{#69047E}{54} \\ & = \boxed{89} \end{aligned}$

This is my own solution:

Take $\alpha = x+y$ and $\beta=xy$ , noting that $x^3+y^3 = \alpha^3-3\alpha\beta$ .

The given equations become $\begin{cases} \alpha\cdot\beta^4 = 810\\ (\alpha\beta)^3 - 3\alpha\beta^4= 945. \end{cases}$

Substituting the first equation into the second, we get that

$(\alpha\beta)^3 = 945+3\cdot 810 = 3375\Longrightarrow \alpha\beta = 15.$

Combining this with the first equation, we find that

$\beta^3 = 810/15 = 54$

and

$\alpha^3 = 3375/54 = 125/2.$

The problem asks us to compute

$2\alpha^3 + \beta^3 - 6\alpha\beta = 125 + 54 - 90 = 89.$

We can factor the equations given to us into $x^4y^4(x+y)=810$

$x^3y^3(x^3+y^3)=945$

The expression we want is the same as $2(x^3+y^3)+(xy)^3=\dfrac{1890}{x^3y^3}+x^3y^3$ . Thus we want the value of $x^3y^3$ . Cubing the first equation gives us

$x^{12}y^{12}(x^3+y^3+3xy(x+y))=810^3 \implies x^{12}y^{12}\left(\dfrac{945}{x^3y^3}+\dfrac{3\cdot810\cdot xy}{x^4y^4}\right)=810^3$

$\implies (945+3\cdot810)x^9y^9=810^3 \implies x^3y^3=\dfrac{810}{15}=54$

Therefore, the answer is $\dfrac{1890}{54}+54=\boxed{89}$

From $x^3y^3(x^3+y^3)=x^6y^3+x^3y^6=945,$ we obtain that $x^3y^3[(x+y)^3-3xy(x+y)]=945$ . So $x^3y^3(x+y)^3=945+3x^4y^4(x+y)=945+3(x^5y^4+x^4y^5)=945+3*810=3375$ . Taking cube root of both sides we get that $xy(x+y)=15$ . Dividing $x^4y^4(x+y)=x^5y^4+x^4y^5=810$ side by side by $xy(x+y)=15$ we obtain $x^3y^3=54\:\:\: (*)$

Now, dividing $x^3y^3(x^3+y^3)=x^6y^3+x^3y^6=945$ side by side by $x^3y^3=54$ we obtain $x^3+y^3=\frac{945}{54}=\frac{35}{2}\:\:\:(**)$ Therefore, using $(*)$ and $(**)$ , $2x^3+(xy)^3+2y^3=2(x^3+y^3)+(xy)^3=2(\frac{35}{2})+54=89.$

Divide it 945/810=(x²-xy+y²)/(xy)=7/6 So 6x²-6xy+6y²=7xy i.e. 6x²-13xy+6y²=0 and divide by y² and replace x/y=t so we'll get 6t²-13t+6=0 i.e. t^(±1)=3/2 So x/y =3/2 ( for 2/3 is similiar ) 2x=3y,so from 32x⁴y⁴(x+y)=32×810 We get (2x)⁴y⁴(2x+2y)=32×810 i.e. (3y)⁴y⁴(3y+2y)=32×81×5×2 so y^9=64=2^6 so y³=4,so (2/3x)³=4, i.e. x³=27/2 so now is our answer 2×4+54+27=89

basically just factor and equate and you get $x^3y^3=54$ and what we're trying to find is $2(\frac{945}{x^3y^3})+x^3y^3$