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We have the following equations:

$x^3 + px + q = 0$

$x = u + v $

$p=-3uv$

Where $q$ and $p$ are known real numbers and $x$ is an unknown real number. We want to find an equation for $x$ by first writing one for $u$ and $v$. So we get:

$$ (u+v^3) - (u+v)3uv +q = 0$$

$$ u^3 + v^3 = -q$$

$$ (u+v)(u^2-uv+v^2) = -q$$

$$ x((u+v)^2 - 3uv) = -q $$

$$ x(x^2+p) = -q $$

But after this I hit a brick wall. I can't seem to isolate the $x$ for the life of me, where did I go wrong?

Community
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Preet Singh
  • 73
  • You'll notice that you've arrived back at your original equation! :) Try combining $u^3 + v^3 = - q$ with the fact that $p^3 = - 27 u^3 v^3$. (To reduce exponent clutter, you can think of this as $U+V=-q$ and $p^3 = -27 UV$. Solve for $U$ and $V$.) – Blue Sep 17 '14 at 21:54
  • http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method – Will Jagy Sep 17 '14 at 22:45

1 Answers1

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HINT:

You have two equations:

$u^3 + v^3 = -q,\,\,p=-3uv$

georg
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