# Cubic Equations

• August 10th 2011, 01:23 AM
maloneb
Cubic Equations
I have a cubic equation that I need to find a real solution to, does anyone have any ideas??

The equation is ((A/B)-1)*(((A/C)^2)-1)=2*(A/C)*((D*E)/F)

where -1 is actually -1 and not the inverse. All values are known except for that of A, this is the value that I need to find.

Cheers,

Ben
• August 10th 2011, 01:26 AM
Duke
Re: Cubic Equations
rearange so you have F(a)=0, multiplying through by any denominators. If the values are known, can't you use those instead of B,C,D,E,F
• August 10th 2011, 01:30 AM
maloneb
Re: Cubic Equations
Well the issue is that I'm rearranging the formula to use in a VBA program, so I need to input the variables as B,C,D,etc as the values are calculated earlier in the program. So I need an equation which solves A in terms of B,C,D,etc. I'm just having an absolute nightmare in reformulating it. Keep coming to dead ends.

(The value of F begins low and then increases until A is over a certain value)
• August 10th 2011, 06:36 AM
HallsofIvy
Re: Cubic Equations
If those coefficients can be anything, then you will need the general "cubic" formula.

Let a and b be any two numbers. $(a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$
Also, $3ab(a+ b)= 3a^2b+ 3ab^3$ so subtracting, we have $(a+ b)^2- 3ab(a+ b)= a^3+ b^3$

If we let x= a+ b, m= 3ab, and $n= a^3+ b^3$, that equation is $x^3- mx= n$.

Now, the question is, if we know m and n, can we solve for a and b- and so find x?

From m= 3ab, we get b= m/3a. Putting that into $a^3+ b^3= n$, we have $a^3+ \frac{m^3}{3^3a^3}= n$. Multiply both sides by $a^3$ to get $(a^3)^2+ \frac{m^3}{3^3}= na^3$ or $(a^3)^2- na^3+ \frac{m^3}{3^3}= 0$ which we can think of as a quadratic equation for $a^3$.

By the quadratic formula, $a^3= \frac{n\pm\sqrt{n^2- 4\left(\frac{m}{3}\right)^3}}{2}$.
We can take that "2" in the denominator into the root and write
$a^3= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^2}$

Because $a^3+ b^3= n$, $b^3= n- a^3= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^2}$

That can be used to solve the "reduced cubic" with no " $x^2$" term.

For the general cubic equation, $ax^3+ bx^2+ cx+ d= 0$, first let y= x+ q, for some unknown number q, so that
$ax^3+ bx^2+ cx+ d= a(y- q)^3+ b(y- q)^2+ c(y- q)+ d$. Multiply that out and choose q so that the coefficient of $y^2$ is 0.