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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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\(\displaystyle (z-y)^3-3(z-y)^2+5(z-y)-17=0\)

Which becomes:

\(\displaystyle z^3-3yz^2+3y^2z-y^3-3z^2+6yz-3y^2-5y+5z-17=0\)

\(\displaystyle z^3-3yz^2+3y^2z-3z^2+6yz-3y^2+5z-6-\left(y^3+5y+11 \right)=0\)

Using the second equation, this becomes:

(1) \(\displaystyle z^3-3yz^2+3y^2z-3z^2+6yz-6y^2+5z-6=0\)

Now, the second equation may be written:

\(\displaystyle (z-x)^3-3(z-x)^2+5(z-x)+11=0\)

Which becomes:

\(\displaystyle z^3-3xz^2+3x^2z-x^3-3z^2+6xz-3x^2-5x+5z+11=0\)

\(\displaystyle z^3-3xz^2+3x^2z-3z^2+6xz-3x^2+5z-6-\left(x^3+5x-17 \right)=0\)

Using the first equation, this becomes:

(2) \(\displaystyle z^3-3xz^2+3x^2z-3z^2+6xz-6x^2+5z-6=0\)

Adding (1) and (2), and simplifying, we have:

\(\displaystyle (z-2)\left(3\left(x^2+y^2 \right)-\left(z^2+2z-6 \right) \right)=0\)

Hence:

\(\displaystyle z=x+y=2\)

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- Feb 14, 2012

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I just love your approach so so much!

- Jan 26, 2012

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$u^3+2u-14=0$ and $v^3+2v+14=0$.

Adding these two gives

$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.

Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.

- Aug 30, 2012

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(Shakes his head.) Wow!

$u^3+2u-14=0$ and $v^3+2v+14=0$.

Adding these two gives

$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.

Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.

-Dan

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- Feb 14, 2012

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Thanks for participating,

$u^3+2u-14=0$ and $v^3+2v+14=0$.

Adding these two gives

$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.

Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.