I have this string of questions, but when I look up depressed cubic, I don't quite understand. I'm not asking for all of these questions to be answered explicitly, but for some explanation on depressed cubics and solving them algebraically. Thanks.
a) Show that the βdepressedβ cubic equation $ππ₯^3 + ππ₯ + π = 0$ can be solved geometrically for its real roots on a rectangular Cartesian coordinate system on which the cubic curve $π¦ = π₯^3$ has been carefully drawn, by merely drawing the line $ππ¦ + ππ₯ + π = 0$. Explain.
b) Solve, by the method of (a), the cubic equation $π₯^3 + 6π₯ β 15 = 0$.
c) Solve the cubic equation $4π₯^3 β 39π₯ + 35 = 0$ geometrically.
d) Show that any complete cubic equation $ππ₯^3 + ππ₯^2 + ππ₯ + π = 0$ can be reduced to the βdepressedβ form in the variable π§ by the substitution: $π₯ = π§ β \frac{b}{3a}$.
e) Now solve the cubic equation $π₯^3 + 9π₯^2 + 20π₯ + 12 = 0$ geometrically as described in (a) and (d).
Best Answer
Well, a depressed cubic as you put it; has no quadratic part (no $x^2$ term). The trick here is to realise that when you are trying to find its root, you are solving the equation:
$$ ax^3 + bx + c = 0 $$
You'll see that if you substitute:
$$ y = x^3 $$
you'll just get:
$$ a(x^3) +bx + c= ay+bx+c = 0 $$
So it amounts to solving $ay+bx+c = 0$ simultaneously with $x^3$. You can see this in this desmos demonstration:
https://www.desmos.com/calculator/xkcdxf7sy2
ps: this is the basic idea on which the cubic equation is derived ;)