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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 “depressed” cubic equation $^3 + + = 0$ can be solved geometrically for its real roots on a rectangular Cartesian coordinate system on which the cubic curve $ = ^3$ has been carefully drawn, by merely drawing the line $ + + = 0$. Explain.

b) Solve, by the method of (a), the cubic equation $^3 + 6 − 15 = 0$.

c) Solve the cubic equation $4^3 − 39 + 35 = 0$ geometrically.

d) Show that any complete cubic equation $^3 + ^2 + + = 0$ can be reduced to the “depressed” form in the variable by the substitution: $ = − \frac{b}{3a}$.

e) Now solve the cubic equation $^3 + 9^2 + 20 + 12 = 0$ geometrically as described in (a) and (d).

Jyrki Lahtonen
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K Math
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  • More interesting is the case of intersecting a circle with a cubic curve. This is a sixth degree system, but only certain sextic equation -- and therefore, only certain quintic equations -- can be solved this way. E.g. we can set up a solution for constructing a regular hendecagon, but not for quintisecting an angle. – Oscar Lanzi Sep 01 '21 at 12:29

3 Answers3

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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 ;)

user2662833
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A depressed cubic equation is a cubic with the second degree term missing. (i.e. b=0) Every cubic equation can be transformed into a depressed cubic using the substitution in d).

Do you agree that the curve $y=x^3$ and the line $ay+bx+c=0$ will intersect at the point whose $x$ coordinate satisfies the equation $ax^3+bx+c=0$?

You are not being asked to solve algebraically at this point! This was only the first step in solving algebraically. Then your reduced equation (with a=1) is compared with the identity $$(e+f)^3-3ef(e+f) -(e^3+f^3)=0$$ $$x^3+6x-15=0$$ $x$ will be $e+f$, $-3ef=6$ and $e^3+f^3=15$ The second equation lets you find $f$ in terms of $e$ Then substituting into the third equation gives you a quadratic in $e^3$ which you solve. You should then take the cube root and find $f$ and add them together to get $x$.

Teymour
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judith Khan
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  • I'm not sure I understand where you're coming from with your e and f notation? – K Math Oct 11 '18 at 16:05
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    That equation is an identity. It is always true, no matter what letters we use. The deal is we are trying to find x, so we decide to break it into a sum but we don't know what the addends are, Pretending x=e+f and using that identity we can find the $e^3$ and $f^3$ – judith Khan Oct 11 '18 at 16:16
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If you were to graph points and try to solve the general cubic graphically, you would iteratively calculate points until you found the place that the curve crossed zero -- guess and check for every cubic you needed to solve.

To solve it via the depressed cubic insight, you pull out your piece of graph paper on which you have already graphed the equation $y'=x'^3$ and have written (once and for all) the depressed cubic for the generic cubic equation, $y'= Bx'+C$ with $D=b/(3a)$, $B=(Db-c)/a$ and $C=(Dc-d)/a-2D^3$ that was obtained by rewriting $y=ax^3+bx^2+xc+d$ with the substitution $x = x'-D$, $y=0$, and $y'=x^3$ and expanding.

Calculate $B$ and $C$ and pick two points on that line (e.g. solve when $x'=0$ and when $y'=0$).

Connect those two points with a line and see where it crosses the graphed simple cubic.

Subtract $b/(3a)$ from the $x'$ value(s) at the crossing point(s) and that is where the original cubic would crossed the x-axis.

For $x^3+9x^2+20x+12$ you get $0=y'-7x'+6$ and find that the line crosses the simple cubic at {-3,1,2}; subtracting $D=9/(3 \cdot1)=3$ gives the roots of the original equation: {-6,-2,-1}.

smichr
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