How do we derive the so called cubic formula without using Cardano's method or substitution? I would like to see a step by step proof of where wolfram alpha derives this answer. And also explain where the factor of $\dfrac{1\pm i\sqrt{3}}{6\sqrt[3]2a}$ comes from in the 2nd and 3rd roots.
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$\begingroup$ wolframalpha.com/input/?i=ax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0 $\endgroup$– TeocNov 16, 2014 at 21:48
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$\begingroup$ Did you check Wikipedia? This is far more complex than simply "completing a cube". $\endgroup$– user2345215Nov 16, 2014 at 21:48
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$\begingroup$ Completing the cube leads you to a depressed cubic, $x^3+ax+b$, which, unlike its quadratic counterpart $x^2+a$, does not possess a previously-known solution. $\endgroup$– LucianNov 16, 2014 at 21:56
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1 Answer
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A general form for the cubic equation is,
$$ax^3+bx^2+cx+d=0 \tag{1}$$
To find the roots of this equation we first try to get rid of the quadratic term $x^2$. The substitution $x=y-\dfrac{b}{3a}$ helps in achieving our goal. This results in, $$ay^3+\left(c-\dfrac{b^2}{3a}\right)y+\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{2}$$ which we transform into the following, $$y^3+\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)y+\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{3}$$ Upon assuming $e=\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)$ and $f=\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)$ we get the equation as, $$y^3+ey+f=0\tag{4}$$ We reduce this equation by the substitution $y=z+\dfrac{s}{z}$ and choosing $s=-\dfrac{e}{3}$ we obtain the simplified equation as, $$z^6+fz^3-\dfrac{e^3}{27}=0\tag{5}$$ What only remains is to make the substitution $u=z^3$.
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1$\begingroup$ And for the factor: $(1\pm i \sqrt{3})/2$ ... you get this when you take the cube root of a complex number. In the above, taking solution $u$ and converting to solution $z$. $\endgroup$– GEdgarNov 26, 2014 at 14:04
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1$\begingroup$ This is truly something to behold. Thanks for this answer. $\endgroup$ Jul 14, 2015 at 17:29
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$\begingroup$ In fact what we do is to investigate the quadratic $z^2-yz+s=0$ where we treat $y$ as a constant. Here clearly the roots are of the form $\alpha$ and $\frac{\alpha}{s}$. $\endgroup$– user170039Sep 13, 2015 at 3:05
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1$\begingroup$ This was so much easier than I expected. @TheTurtle I'm sure there are other methods to derive the cubic formula, which avoid division by $0$. $\endgroup$ May 20, 2016 at 21:51