MHB Find the sum of a^(1/3)+b^(1/3)+c^(1/3)

[anemone]

Let $a,\,b$ and $c$ be real numbers such that

$a+b+c=ab+bc+ac=-\dfrac{1}{2}\\abc=\dfrac{1}{8}$

Evaluate $a^{\tiny\dfrac{1}{3}}+b^{\tiny\dfrac{1}{3}}+c^{\tiny\dfrac{1}{3}}$.

 

[Opalg]

Let $a,\,b$ and $c$ be real numbers such that

$a+b+c=ab+bc+ac=-\dfrac{1}{2}\\abc=\dfrac{1}{8}$

Evaluate $a^{\tiny\dfrac{1}{3}}+b^{\tiny\dfrac{1}{3}}+c^{\tiny\dfrac{1}{3}}$.

[sp]The equation with roots $a$, $b$ and $c$ is $x^3 + \frac12x^2 - \frac12x - \frac18 = 0$.

Let $y^3 - sy^2 + ty - \frac12=0$ be the equation with roots $a^{1/3}$, $b^{1/3}$ and $c^{1/3}$. (Note that the constant term is $-(abc)^{1/3} = -\frac12$.) The sum of the roots of that equation is $s = a^{1/3} + b^{1/3} + c^{1/3}$, so we want to find $s$.

Put $y^3=x$ to see that the equation $x - sx^{2/3} + tx^{1/3} - \frac12=0$ has roots $a$, $b$ and $c$. Write that equation as $x - \frac12 = sx^{2/3} - tx^{1/3}$, then cube both sides to get $$\bigl(x - \tfrac12\bigr)^3 = x(sx^{1/3} - t)^3,$$ $$\begin{aligned} x^3 - \tfrac32x^2 + \tfrac34x -\tfrac18 &= x(s^3x - 3s^2tx^{2/3} + 3st^2x^{1/3} - t^3) \\ &= x\bigl(s^3x - 3st(sx^{2/3} - tx^{1/3}) - t^3\bigr) \\ &= x\bigl(s^3x - 3st(x - \tfrac12) - t^3\bigr), \end{aligned} $$ $$x^3 - x^2\bigl(\tfrac32 + s^3 - 3st\bigr) +x\bigl(\tfrac34 - \tfrac32st + t^3\bigr) - \tfrac18 = 0.$$ Compare that with the original equation for $x$ to see that $$s^3 = -2 + 3st,$$ $$ t^3 = -\tfrac54 + \tfrac32st.$$ Multiply those two equations to get $$s^3t^3 = \tfrac52- \tfrac{27}4st + \tfrac92s^2t^2.$$ Now multiply by $8$ and write $z = 2st$: $$z^3 - 9z^2 + 27z - 20 = 0,$$ $$(z-3)^3 = -7,$$ $$z = 3 - \sqrt[3]7.$$ So $ st = \frac12(3 - \sqrt[3]7)$ and therefore $$s^3 = -2 + 3st = \frac{5 - 3\sqrt[3]7}2.$$ Finally, $s = {\large \sqrt[3]{\dfrac{5 - 3\sqrt[3]7}2}} \approx -0.7175.$

[/sp]

 

[anemone]

Thanks Opalg for your awesome solution!(Cool)