Help Please CPhill, Melody, Or heureka

Let a, b, c   be the real roots of   $x^3 -4x^2 -32x+17=0.$

Solve for  x in:                            $\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0.$

Hello Guest!

WolframAlpha calculated:

a = - 4.3195

b = 0.50355

c = 7.8159

a + b + c = 4

abc = - 17

$\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0\\ \sqrt[3]{x +4.3195} + \sqrt[3]{x - 0.50355} + \sqrt[3]{x - 7.8159} = 0$

No solution found.

  !

Hi asinus,

That was good thinking.

But I graphed your equation using Desmos and there was a solution

Hello Guest!

If a,b,c are the solutions of the polynomial equation:    $x^3-4x^2-32x+17=(x-a)(x-b)(x-c)=0$ 

Then, by Vietas' formulae:                   $a+b+c=4$

                                                           $ab+ac+bc=-32$ 

                                                                  $abc=17$

Using the following identity:  

$y_1^3+y_2^3+y_3^3-3y_1y_2y_3=(y_1+y_2+y_3)(y_1^2+y_2^2+y_3^2-y_1y_2-y_1y_3-y_2y_3)$  (*)

(More commonly written as: $x^3+y^3+z^3=3xyz$ if and only if $x+y+z=0$).

Then:

Let  $y_1=\sqrt[3]{x-a}$ , $y_2=\sqrt[3]{x-b}$ , $y_3=\sqrt[3]{x-c}$

We are given: $y_1+y_2+y_3=0$

Hence, (*) becomes: 

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

   (Notice: $(x-a)(x-b)(x-c)=x^3-4x^2-32x+17$)

Thus,

            $3x-(a+b+c) = 3\sqrt[3]{x^3-4x^2-32x+17}$   

            (By Vietas formula: a+b+c=4, substitute and cube both sides)

$\iff (3x-4)^3=27(x^3-4x^2-32x+17)$

Expanding:

$\iff 27x^3-108x^2+144x-64=27x^3-108x^2-864x+459$

Simplify to get a linear equation:

$144x-64=-864x+459 \implies 1008x=523 \implies x=\frac{523}{1008}$

Therefore,  $x=\dfrac{523}{1008}$

Hi asinus, by the way, Wolframalpha approximated a,b,c a lot.

Because when using mathway.com, it gives the following values: 

$a=-4.31946756$

$b=0.5035452$

$c=7.81592235$

So:

$\sqrt[3]{(x+4.31946756)}+\sqrt[3]{(x-0.5035452)}+\sqrt[3]{(x-7.81592235)}=0$

And, then graphing this equation as Melody did (But using the more accurate values of a,b,c):

Giving the approximation: $0.519$   (And the exact solution is: $\frac{523}{1008}=0.518849206..$).

Hope this helps!