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I am trying to solve $$ x^3+7x^2+15x+15=0 $$. The real solution was not an issue. However, the complex solutions driving me mad.

I was able to express the real solution as $$ x=(1/3) \left (-7+\sqrt[3]{-73+9\sqrt{65}}+\sqrt[3]{-73-9\sqrt{65}} \right ). $$ I managed to manipulate the cube roots as $$ \sqrt[3]{-73+9\sqrt{65}}=\frac{\sqrt[3]{73-9\sqrt{65}}+i\sqrt{3}\sqrt[3]{73-9\sqrt{65}}}{2} $$ and $$\sqrt[3]{-73-9\sqrt{65}}=\frac{\sqrt[3]{73+9\sqrt{65}}+i\sqrt{3}\sqrt[3]{73+9\sqrt{65}}}{2}. $$

Substituting these complex values for the cube roots in my original expression for $x$ yields the correct real part, but the incorrect imaginary part. I do know that if I simply move the $ \sqrt{3} $ from the numerator to the denominator, then it does yield the correct imaginary part. I am at a complete loss as to how this shift occurs. Any help would be greatly appreciated!

Blue
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Distilled Ink
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  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Dec 23 '24 at 16:02
  • @Distilled Ink: The usual reduction for disappear the second degree term can be simplificated by the form $(3x+7)^3-12(3x+7)+146$ so you can solve the standard way the equation $X^3-12X+146=0$ where it is clear what $X$ is. – Ataulfo Dec 23 '24 at 16:39
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    Are you remembering that each cube root has three possible values? – Ted Shifrin Dec 23 '24 at 16:40
  • @Piquito I absolutely should be approaching it that way. However, I am stubbornly interested in how the complex roots pop out of the general cubic formula expressed in terms of a, b, c, and d. I'm in too deep to quit now. – Distilled Ink Dec 23 '24 at 16:58
  • @Ted Chifrin I have been thinking about that, but clearly not in the correct way. Perhaps another cup of coffee will help. Thank you! – Distilled Ink Dec 23 '24 at 16:59
  • Perhaps https://math.stackexchange.com/q/2838797/349785 helps. – Paul Frost Dec 23 '24 at 17:03

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For the real root, it will be simpler to use the hyperbolic solution to obtain $$x_1=-\frac{1}{3} \left(7+4 \cosh\left(\frac{1}{3} \cosh ^{-1}\left(\frac{73}{8}\right)\right)\right)$$

Writing $$x^3+7x^2+15x+15=(x-x_1)(x^2+b x+c)$$ leads to $$b=x_1+7\qquad \text{and} \qquad c=-\frac {15}{x_1}$$

Solve the quadratic equation.

Claude Leibovici
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