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- #1

- Feb 14, 2012

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$x^3=3x-12y+50\\y^3=12y+3z-2\\z^3=27z+27x$

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,587

$x^3=3x-12y+50\\y^3=12y+3z-2\\z^3=27z+27x$

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- #2

- Feb 14, 2012

- 3,587

$x^3-3x-2=(x-2)(x+1)^2\\y^3-12y-16=(y-4)(y+2)^2\\z^3-27z-54=(z-6)(z+3)^3$

Suppose $x>2$, we then have

$-12y+50=x^3-3x>2\implies y<4$

$z^3-27z=27x>54 \implies z>6$

$y^3-12y=3z-2>16 \implies y>4$

which leads to a contradiction.

Now, assume $x<2$, we then have

$-12y+50=x^3-3x<2 \implies y>4$

$3z-2=y^3-12y>16 \implies z>6$

But this leads to

$27x=z^3-27z>54$ which is impossible.

THus, $x=2$ and that gives the only solution set $(x,\,y,\,z)=(2,\,4,\,6)$.