Omar Khayyam's method for solving cubics So I need to answer the following question using Khayyam's method.  I can get the answer using modern methods, and I know the basics of his method, but I cannot figure out how to find the two conic sections in order to find their intersection for the solution.  Here's the question:
Set up the solution for the following cubics using the method of Omar Khayyam by identifying the solution as an intersection of two explicitly given conic sections.  For example, the cubic equation $x^3 = 2$ can be solved by finding the intersection of the parabola $y = x^2$ and the parabola $y^2 = 2x$.  
(a)  $x^3 = 9x + 18$
(b)  $x^3 + 9x = 18$
Thanks so much.
Thanks for the answer, guys.  I needed that equation for finding the two equations.  Is there a similar equation that can be used for part (A)?
 A: In modern terminology, Khayyam solves the cubic $x^3+cx=d$ (your case (b)) as the $x$-coordinate of the point of intersection in the first quadrant between the parabola $x^2=\sqrt{c}y$ and the circle $x\big({d\over c}-x\big)=y^2$. (The center of the circle is at $\bigl({c\over 2d},0\bigr)$, and it passes through the origin $(0,0)$.)
He comments that this gives a unique positive $x$-value, and he shows that this $x$-value solves the cubic equation since 
$${c\over x^2}={x^2\over y^2}={x^2\over x\bigl({d\over c}-x\bigr)}={x\over {d\over c}-x}$$
which implies that $d-cx=x^3$, or $x^3+cx=d$.
Edit: For your case (a), I haven't seen exactly how Khayyam did this. But looking at the above, with the equation $x^3=cx+d$, it would be reasonable to try the same parabola $x^2=\sqrt{c}y$ and the equation
$${c\over x^2}={x^2\over y^2}={x^2\over x\bigl(x+{d\over c}\bigr)}={x\over x+{d\over c}}$$
which, by comparison of the outermost terms, implies that $x^3=cx+d$. For the middle equality to hold, you need the equation $y^2=x^2+{d\over c}x$, which is the equation of a hyperbola with center at $\bigl(-{c\over 2d},0\bigr)$, and which passes through the point $(0,0)$.
Also note that Khayyam of course did not have our algebraic notation at his disposal, so he would have to carry out his arguments using only geometric properties of the conic sections, as developed by Apollonius of Perga.
A: Khayyam Solved much types of $3$rd degree polynomials by geometrically constructing conic sections and their intersections.

http://en.wikipedia.org/wiki/Conic_section#Omar_Khayy.C3.A1m

It will helps you for a particular type of $3$rd degree polynomials :

http://www.math.ntnu.no/~hanche/blog/khayyam.pdf

A: Your example of $x^3 = 2$ could also be expressed as the intersection of $x^2 = y$ and $xy = 2$.
Defining $y = x^2$ makes for an easy way to reduce powers of $x$ to polynomials of smaller degree. You can even reduce quartic polynomials in $x$ to ones of total degree 2 in $x$ and $y$.
