Several different solutions are possible. I managed to find Khayyam's paper online and here are his solutions. Of course he used a purely geometrical language and gave his solution as a construction, followed by a proof that this was indeed right. I'll reverse his reasoning, as you seem to prefer, and use a more familiar language.
First case: $x^3=ax^2+bx+c$.
We can rewrite this equation as:
$$
x^2(x-a)=b\left(x+{c\over b}\right)
$$
and multiply both sides by $(x+{c/ b})$, obtaining:
$$
\tag{1}
x^2(x-a)\left(x+{c\over b}\right)=b\left(x+{c\over b}\right)^2.
$$
If we set now:
$$
\tag{2}
y^2=(x-a)\left(x+{c\over b}\right),
$$
which is the equation of a rectangular hyperbola,
we can rewrite $(1)$ as:
$$
x^2y^2=\left(\sqrt{b}x+{c\over \sqrt{b}}\right)^2.
$$
But this is the square of
$$
\tag{3}
xy=\sqrt{b}x+{c\over \sqrt{b}},
$$
which is the equation of another rectangular hyperbola.
Our equation is then equivalent to the system formed by $(2)$ and $(3)$.
Second case: $x^3+bx=ax^2+c$.
Set:
$$
b=h^2,\quad c=sh^2
$$
and rewrite the equation as
$$
h^2(x-s)=x^2(a-x).
$$
Now multiply both sides by $(x-s)$ (this will introduce the spurious solution $x=s$, which must be discarded) to obtain:
$$
\tag{4}
h^2(x-s)^2=x^2(a-x)(x-s).
$$
If we set now:
$$
\tag{5}
(a-x)(x-s)=(h-y)^2,
$$
which is the equation of a circle, and substitute into $(4)$, we get
$$
h^2(x-s)^2=x^2(h-y)^2.
$$
But this is the square of:
$$
\tag{6}
h(x-s)=x(h-y),
$$
which is the equation of a rectangular hyperbola.
The solution of the original equation can then be obtained as the abscissa of the second intersection between circle $(5)$ and hyperbola $(6)$, their first intersection being point $(s,h)$.
EDIT.
For the case $x^3+ax^2+bx=c$ Khayyam makes the same construction as in the case $x^3+bx=ax^2+c$ outlined above. Just make the substitution $a\to-a$ in the above explanation to find the corresponding hyperbola and circle.
