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If both $x$ and $y$ are rational and nonzero, then $r=\tfrac{x}{y}$ is rational and nonzero. Since $6y^3 = 3x^3 + xy^3$, we deduce that $x = 6 - 3r^3$, and of course $y = \frac{x}{r} = \frac{6-3r^3}{r}$. From the first equation we deduce that
$\begin{array}{rcl}
2x^3 & = & x^2y^4 + 9y^5 \\
2x^3 & = & x^2 \frac{x^4}{r^4} + 9\frac{x^5}{r^5} \; = \; \frac{x^6}{r^4} + \frac{9x^5}{r^5} \\
2 & = & \frac{x^3}{r^4} + \frac{9x^2}{r^5} \\
2r^5 & = & rx^3 + 9x^2 \; = \; 27r(2 - r^3)^3 + 81(2-r^3)^2 \end{array}$
and hence
$f(r) \; =\; 27r^{10} - 162r^7 - 81r^6 + 2r^5 + 324r^4 + 324r^3 - 216r - 324 \; = \; 0$
If we write $r = \frac{a}{b}$ where $a,b$ are coprime integers with $b \ge 1$, then
$27a^{10} - 162a^7b^3 - 81a^6b^4 + 2a^5b^5 + 324a^4b^6 + 324a^3b^7 - 216ab^9 - 324b^{10} \; = \; 0$
Thus we deduce that $b^3$ divides $27a^{10}$, and hence $b^3$ divides $27$, and hence $b = 1,3$. Moreover $a$ divides $324b^{10}$, and so $a$ divides $324$.

If $b=3$ then, since $324 = 4\times81$ we deduce that $a$ divides $4$. Thus $r$ must be one of $\pm\tfrac13,\pm\tfrac23,\pm\tfrac43$. Since none of these six numbers is a zero of $f$, this case does not work.

If $b=1$ then $r=a$ and $f(a)=0$. Since $2a^5 = 27a(2-a^3)^3 + 81(2-a^3)^2$, we deduce that $3$ divides $a$. If $3^2$ divided $a$, then $3^5$ would divide $f(a) + 324 = 324$, which it does not. Thus $r = a$ must be one of $\pm3, \pm6,\pm12$. Since none of these six numbers is a zero of $f$, this case does not work either.

There are no nonzero rational roots to these simultaneous equations.

what do ypu want exactely, to solve problem ou to do something, like jocking
how many solotion do you want
wat is the problem please, i need more details for discussing with you
i am from morroco
nice to know and meet you

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## Comments

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TopNewestthere are nonzero solutions, please give the real problem

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there are nonzero, I am waiting for the real problem

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If both $x$ and $y$ are rational and nonzero, then $r=\tfrac{x}{y}$ is rational and nonzero. Since $6y^3 = 3x^3 + xy^3$, we deduce that $x = 6 - 3r^3$, and of course $y = \frac{x}{r} = \frac{6-3r^3}{r}$. From the first equation we deduce that $\begin{array}{rcl} 2x^3 & = & x^2y^4 + 9y^5 \\ 2x^3 & = & x^2 \frac{x^4}{r^4} + 9\frac{x^5}{r^5} \; = \; \frac{x^6}{r^4} + \frac{9x^5}{r^5} \\ 2 & = & \frac{x^3}{r^4} + \frac{9x^2}{r^5} \\ 2r^5 & = & rx^3 + 9x^2 \; = \; 27r(2 - r^3)^3 + 81(2-r^3)^2 \end{array}$ and hence $f(r) \; =\; 27r^{10} - 162r^7 - 81r^6 + 2r^5 + 324r^4 + 324r^3 - 216r - 324 \; = \; 0$ If we write $r = \frac{a}{b}$ where $a,b$ are coprime integers with $b \ge 1$, then $27a^{10} - 162a^7b^3 - 81a^6b^4 + 2a^5b^5 + 324a^4b^6 + 324a^3b^7 - 216ab^9 - 324b^{10} \; = \; 0$ Thus we deduce that $b^3$ divides $27a^{10}$, and hence $b^3$ divides $27$, and hence $b = 1,3$. Moreover $a$ divides $324b^{10}$, and so $a$ divides $324$.

If $b=3$ then, since $324 = 4\times81$ we deduce that $a$ divides $4$. Thus $r$ must be one of $\pm\tfrac13,\pm\tfrac23,\pm\tfrac43$. Since none of these six numbers is a zero of $f$, this case does not work.

If $b=1$ then $r=a$ and $f(a)=0$. Since $2a^5 = 27a(2-a^3)^3 + 81(2-a^3)^2$, we deduce that $3$ divides $a$. If $3^2$ divided $a$, then $3^5$ would divide $f(a) + 324 = 324$, which it does not. Thus $r = a$ must be one of $\pm3, \pm6,\pm12$. Since none of these six numbers is a zero of $f$, this case does not work either.

There are no nonzero rational roots to these simultaneous equations.

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wow that was fabulos can i meet u in fb

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'Thus we deduce that $b^3$ divides $27a^{10}$ ', Can you explain this to me..?? I didnt get this part. Thanks for the solution BTW.. :)

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$b^3$ is a factor of all the other terms.

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could you please post the answer to this question,kevin

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'Find x and y so that they are rational'.. but isnt Diophantine Equation about integer solutions?

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Thank you for bringing this up I may need to clarify this problem.

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You're Welcome.. but this is a good question anyways.. I'm still working on it! :P

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what do ypu want exactely, to solve problem ou to do something, like jocking how many solotion do you want wat is the problem please, i need more details for discussing with you i am from morroco nice to know and meet you

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Put x as ky and solve

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But that would still mean we have 2 variables.

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Also, k will be a rational number to maintain generality, possibly complicating the question.

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Lol, $x=y=0$ :P

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Lol sorry :D x and y≠0

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