# Solving Cubic Functions

by TheDominis
Tags: cubic, functions, solving
 P: 2 x3 - 12x + 1 = 0 How does one solve for x?
 Mentor P: 20,431 It's not a trivial process. See this Wikipedia article, http://en.wikipedia.org/wiki/Cubic_function especially the Summary about halfway down the page.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 Let x= a- b. Then $a^3= (a-b)^3= a^3- 3a^2b+ 3ab^2- b^3$. Also $3abx= 3ab(a- b)= 3a^2b- 3ab^2$. So $x^3+ 3abx= a^3- b^3$. Letting m= 3ab and $n= a^3- b^3$, then x= a-b satisfies $x^3+ mx= n$. Suppose we know m and n- can we "recover" a and b and so find x? If m= 3ab, then b= m/3a and $n= a^3- m^3/3^3a^3$. Multiplying through by $a^3$ we get $na^3= (a^3)^2- m^3/3^3$ which we can think of as a quadratic equation for $a^3$: $(a^3)^2- na^3- m^3/3^3= 0$ and solve by the quadratic formula: $$a^3= \frac{n\pm\sqrt{n^2+ 4\frac{m^3}{m^3}}}{2}$$$$= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$ so that $$a= \sqrt[3]{\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}}}$$ Since $a^3- b^3= n$, $b^3= a^3- n$ so $$b^3= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$ and $$b= -\sqrt[3]{\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}}}$$ and, of course, x= a- b.

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