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We have, $x^4y^4(x+y)=810$ and
$x^3y^3(x^3+y^3)=945$ or $x^3y^3 (x+y)(x^2-xy+y^2)=945$

On division, $\frac {x^2-xy+y^2}{xy}=\frac {945}{810}=\frac {7}{6}$

$\Rightarrow \frac {x}{y}-1+\frac {y}{x}=\frac {7}{6}$

Let, $\frac {y}{x}=v$

So, $\frac {1}{v}-1+v=\frac {7}{6}$

$\Rightarrow v+\frac {1}{v}=\frac {7}{6}+1=\frac {13}{6}$

Thus, $6v^2+6=13v$

$\Rightarrow 6v^2-4v-9v+6=0$

$\Rightarrow (3v-2)(2v-3)=0$

$\Rightarrow v=\frac {3}{2},\frac {2}{3}$

Since, $v=\frac {y}{x}=\frac {3}{2}$, y=3k & x=2k

So, $(2k)^4.(3k)^4.(2k+3k)=810$

$\Rightarrow 8k^9 =1$

$\Rightarrow k=\frac {1}{2^{1/3}}$

Now, $y=3k=3.2^{-1/3}$ and $x=2k=2.2^{-1/3}=2^{2/3}$

When $v=\frac {2}{3}$, it can be shown that the values of x and y would interchange, that is $x=3.2^{-1/3}$ and $y=2^{2/3}$.

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