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!

Answer 1

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.

Answer 2

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