Just wondering if anyone can clarify these two ways of solving cubics. Theres not many examples online which I can find.
Q1. Give Omar Khayyam’s geometrical solution to the cubic equation $x^3 + 64x = 384.$ Check that the geometric solution is correct using modern methods.
Q2. Then Using the formula established by the 16th-century Italian method, find a solution to the cubic equation $ x^3 + 63x = 316.$
For Q1 I know that we transform this into finding the intersection of a parabola and a hyperbola, however I'm having difficulty applying this formula to an actual example.
I'd appreciate any guidance or examples which can be given
First: Omar Khayyam's method. Let's solve: $$x^3+64x=384$$ We need to turn our equation into something of the form: $$\left(\frac{x^2}{c}\right)^2=x(d-x)$$ With $c$ and $d$ constants. Then we can find the intersection of: $$y^2=x(d-x)$$ $$y=\frac1cx^2$$ The $x$-coordinate of the intersection will be a solution to our equation. Ok, so let's do it: \begin{align} x^3+64x&=384\\ x^3&=384-64x\\ \frac{x^3}{64}&=6-x\\ \frac{x^4}{64}&=x(6-x)\\ \left(\frac{x^2}{8}\right)^2&=x(6-x) \end{align} So, we need to find the intersection of: $$y=\frac18x^2$$ and: $$y^2=x(6-x)$$
Now for Cardano's (The Italian Mathematician) Formula. We have: $$x^3+63x=316$$ $$x^3+63x-316=0$$ Cardano's formula tells us that the equation: $$x^3+px+q$$ has the solutions: $$x=\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}$$ So filling it all in gives: $$x=\sqrt[3]{\frac {316}2+\sqrt{\frac{316^2}4+\frac{63^3}{27}}}+\sqrt[3]{\frac {316}2-\sqrt{\frac{316^2}4+\frac{63^3}{27}}}$$