Algebraic Manipulation with Cube Roots

Question

Let $a,$ $b,$ $c$ be the real roots of $x^3 - 4x^2 - 32x + 17 = 0.$ Solve for $x$ in $$\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0.$$

We probably have to manipulate the $\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c}$ into something more convenient, so we can actually use it to solve the problem. The first thing I tried was cubing the equation; I quit midway through realizing it was a bad idea (It was really messy.)

Next, I got the stupid idea of trying to actually solve the cubic. I got nowhere. (What I was hoping for were some nice solutions for $x.$)

Now, I'm stuck. Help? Thanks!

Possibly use a well known (in contest math, at least) factorization of $A^3 + B^3 + C^3 - 3ABC,$ such as is done in Answer 1 and in Answer 2. — Dave L. Renfro, Sep 06 '20 at 19:16
@Fleccerd I see that you proposed many questions in the past, but never accepted an answer — Vincenzo Tibullo, Sep 07 '20 at 04:41
@Fleccerd how can you say that? It is on your profile page, everybody can see. I respect your will to not accept answers, but please don't deny evidence — Vincenzo Tibullo, Sep 07 '20 at 05:28
@Fleccerd In the remote case you are not aware, by "accepting" the answer we mean clicking on the "tick / check mark" next to an answer so that it becomes green. This is used to mark one of the answers as the "accepted" one, the one which in your opinion answers the question better. You can only accept one answer for each question. You seem to have never done this in the past, according to your profile. If you browse the site, you'll notice these green ticks on many questions (not all, but most). — chi, Sep 07 '20 at 07:53

Vincenzo Tibullo · Accepted Answer · 2020-09-07T06:45:27.710

11

Elevating to third power we get $$ 3 \left[x+\left(\sqrt[3]{x-a}+\sqrt[3]{x-b}\right) \left(\sqrt[3]{x-a}+\sqrt[3]{x-c}\right) \left(\sqrt[3]{x-b}+\sqrt[3]{x-c}\right)\right]-(a+b+c)=0 $$ using the original equation gives $$ 3 \left[x-\sqrt[3]{x-a}\sqrt[3]{x-b}\sqrt[3]{x-c}\right]-(a+b+c)=0\\ 3 \left[x-\sqrt[3]{(x-a)(x-b)(x-c)}\right]-(a+b+c)=0\\ $$ but $$ a+b+c=4\\ (x-a)(x-b)(x-c)=x^3-4x^2-32x+17 $$ so the equation becomes $$ 3 \left[x-\sqrt[3]{x^3-4x^2-32x+17}\right]-4=0\\ $$ and isolating the cube root and elevating again to the third power $$ x^3-4x^2-32x+17=\left(x-\frac{4}{3}\right)^3\\ $$ simplifying we get $$ 1008x=523\quad\implies\quad x_0=\frac{523}{1008} $$ Be aware that the cube root is intended to be defined on the whole $\mathbb{R},$ otherwise no real solution would exist, given that one of $x_0-a,x_0-b,x_0-c$ is negative.

edited Sep 07 '20 at 06:45

answered Sep 06 '20 at 19:35

Vincenzo Tibullo

10,955

Very good..... +1 – Sebastiano Sep 06 '20 at 20:41
OMG THAT IS SO CLEVER
thank you so much!
– Fleccerd Sep 07 '20 at 05:22
I have attempted to give a different solution, accompanied by graphical representations that help (so I think) in the understanding of this very special question... – Jean Marie Oct 18 '23 at 12:40

score 0 · Answer 2 · answered Oct 18 '23 at 12:36

I would like to say that there exists a slightly less "magic wand" solution.

Let $p(x)=x^3 - 4x^2 - 32x + 17.$

We will use later on the fact that the sum of its roots is $4$ (one of Vieta's formulas).

Let us give the names $g(x), A, B, C$ in the following expression :

$$g(x)=\underbrace{\sqrt[3]{x-a}}_A+\underbrace{\sqrt[3]{x-b}}_B+\underbrace{\sqrt[3]{x-c}}_C$$

Our hypothesis is that :

$$A+B+C=0\tag{1}$$

Now, consider classical identity (as advised by @Dave L. Renfro) :

$$A^3+B^3+C^3-3ABC=(A+B+C)\tfrac12\underbrace{((A-B)^2+(B-C)^2+(C-A)^2)}_{E}$$

Identity (1) implies

$$A^3+B^3+C^3-3ABC=0$$

Otherwise said :

$$(x-a)+(x-b)+(x-c)=3\sqrt[3]{(x-a)(x-b)(x-c)}$$

$$3x-(a+b+c)=3\sqrt[3]{p(x)}$$

$$(x-\tfrac43)^3=p(x)$$

Expanding the LHS, a big simplification occurs, leaving a first degree equation... whose root $x=\frac{523}{1008}\approx 0.51885$ is the looked-for value.

Remark : I have said upwards : "identity (1) implies" ; in fact a closer look shows that it is an "equivalence" because expression $E$ cannot be zero unless $A=B=C$ which cannot be the case.

Fig. 1 : On the same figure, a plot of functions $p$ (in fact $\frac{1}{20}p$) and $q$. Please note the inflection points of $q$ whose abscissas are the roots $\approx -4.3194676,\ \ 0.50354521,\ \ 7.8159224$ of $f$.

SAGE program for the generation of the figure (its interest is to show how the cubic root function $r3$ has to be re-defined)

var('x')
def r3(x) :
  if x > 0:
  y = (x)^(1/3)
else:
  y = -(-x)^(1/3)
return y
#s=solve(x^3 - 4*x^2 - 32*x + 17==0,x)
def g(x) :
  z = r3(x-0.50354521)+r3(x+4.3194676)+r3(x-7.8159224)
return z
G1=plot(0.05*(x^3 - 4*x^2 - 32*x + 17),-6,9,color='red');
G2=plot(g,-6,9);
G1+G2

Algebraic Manipulation with Cube Roots

2 Answers2