Find the value of: $\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$

Question

Let $a,b,c$ be roots of the cubic $$x^3-x^2-2x+1=0$$ Then, find the value of: $$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$$

My attempt.

I used the substitutions $$a+b=x^3, b+c=y^3, a+c=z^3$$

$$x^3+y^3+z^3=2(a+b+c)=2$$

Then I used the identity $$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz$$

But I stuck here. I can not simplify $$xy+yz+xz$$

I may be mixing up the coefficients, but don't you have $a+b+c=-1,ab+bc+ac=2$? — abiessu, Nov 24 '22 at 16:58
In case of interest, this is the minimal polynomial of $-2\cos \frac{2\pi}7$, and belongs to a family of polynomials described here https://math.stackexchange.com/a/1067992/4781 — Macavity, Nov 24 '22 at 18:25

Robert Z · Accepted Answer · 2022-11-24T19:16:48.870

5

From the given cubic, we know that $a+b+c=1$, therefore $$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}=\sqrt[3]{1-c}+\sqrt[3]{1-a}+\sqrt[3]{1-b}=\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$ where $A,B,C$ are the roots of $(1-x)^3-(1-x)^2-2(1-x)+1=0$, that is $x^3-2 x^2-x+1=0$. It can be shown (see for example this link) that $$x^3+(\alpha^3−3\alpha\beta +3\gamma)x^2+ (\beta^3−3\alpha\beta \gamma+3\gamma^2)x+\gamma^3=0$$ is the cubic equation whose roots are the cubes of the roots of $x^3+\alpha x^2+\beta x+\gamma=0$. So it remains to find $-\alpha$ in the system $$\begin{cases} \alpha^3−3\alpha\beta +3\gamma=-2\\ \beta^3−3\alpha\beta \gamma+3\gamma^2=-1\\ \gamma^3=1 \end{cases}$$ which boils down to $\gamma=1$, $\beta=\frac{\alpha^3+5}{3\alpha}$ and $$(\alpha^3+5)^3-27\alpha^3(\alpha^3+5)+4\cdot 27\alpha^3=0$$ or $$\alpha^9-12\alpha^6+48\alpha^3+125=(\alpha^3-4)^3+189=0$$ which has just one real root at $\alpha=-\sqrt[3]{3\sqrt[3]{7}-4}$. Hence $$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}=-\alpha=\sqrt[3]{3\sqrt[3]{7}-4}\approx 1.20249.$$

P.S. Notice that the roots $a,b,c$ of the cubic polynomial $f(x)=x^3-x^2-2x+1$ are three distinct real numbers (one is negative and two are positive) because $f(−2)<0$, $f(0)>0$, $f(1)<0$, $f(2)>0$.

edited Nov 24 '22 at 19:16

answered Nov 24 '22 at 18:05

Robert Z

145,942

Thank you, is it possible to expand your answer? – user1094359 Nov 24 '22 at 18:12
But how do you know that, the answer is real ? – user1094359 Nov 24 '22 at 18:14
"But how do you know that, the answer is real?" Because the given cubic $x^3-x^2-2x+1=0$ has three real roots. – Robert Z Nov 24 '22 at 18:18
Then how do you know that, all roots are real? – user1094359 Nov 24 '22 at 18:21
2

If $f(x)=x^3-x^2-2x+1$ then $f(-2)<0$, $f(0)>0$, $f(1)<0$, $f(2)>0$ then apply the intermediate value theorem. – Robert Z Nov 24 '22 at 18:28
Can you provide an elementary way? – user1094359 Nov 24 '22 at 18:37
Only for high school level. There is no such a theorem in highschool program – user1094359 Nov 24 '22 at 18:42
Are you asking an elementary way for showing that $f(x)=x^3-x^2-2x+1$ has three distinct real roots? Since $f(-2)<0$ and $f(0)>0$, do you agree that $f$ has at least a real root in $(-2,0)$? – Robert Z Nov 24 '22 at 18:45
Yes exactly you are right. – user1094359 Nov 24 '22 at 18:46
@RobertZ High school contest math. – user1094359 Nov 24 '22 at 18:50
@Macavity Thank you. I wish I could understand this more descriptively in the answer. I hope I can understand – user1094359 Nov 24 '22 at 18:53
@user1094359 Maybe my solution is long (I will be happy to read a better one), but I think it is no so advanced. Take your time to understand it. – Robert Z Nov 24 '22 at 18:56
Thanks. Why $\gamma^3=1$ implies $\gamma=1$? Because, we have $3$ complex roots and why you are looking for $-\alpha$, instead of $\alpha$? – user1094359 Nov 24 '22 at 19:03
$\gamma$ is real because the given cubic has all real roots. The coefficient $alpha$ is the opposite of the sum of the roots. – Robert Z Nov 24 '22 at 19:10
See https://math.stackexchange.com/a/4497270/72031 – Paramanand Singh Nov 27 '22 at 23:56

Find the value of: $\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$

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