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# Help Please CPhill, Melody, Or heureka

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

Jun 20, 2022

#1
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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.

!

Jun 20, 2022
#2
+118477
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Hi asinus,

That was good thinking.

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

Jun 21, 2022
#3
+2

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,  $$x=\dfrac{523}{1008}$$

Jun 21, 2022
#4
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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!

Jun 21, 2022