Roots of a Quadratic Problem

Question

I'm struggling with this problem and was hoping I could get some advice. Here is the problem:

Let a and b be the roots of the quadratic equation $x^2−x−1/27=0$.

Without calculating the a and b show that $a^{1/3}+b^{1/3}$ is a root of the equation $x^3+x−1=0$.

Any help would be much appreciated!

Thanks

lab bhattacharjee · Accepted Answer · 2013-10-23T12:43:41.193

4

Using Vieta's formula, $a+b=1$ and $\displaystyle ab=\frac{-27}1$

Let $y=a^{\frac13}+b^{\frac13}$

$$\implies y^3=a+b+3(ab)^{\frac13}(a+b)=1+3(-27)^{-\frac13}y$$

Now, one of three values of $(-27)^{-\frac13}$ is $\{(-3)^3\}^{-\frac13}=-\frac13$

$$\implies y^3=1-y$$

edited Oct 23 '13 at 12:43

answered Oct 23 '13 at 12:37

lab bhattacharjee

274,582

Okay, Thanks! Could you just explain your 3rd step please? I don't quite follow it. – Hummus Oct 23 '13 at 12:42
@Hummus, please have a look into the edited version – lab bhattacharjee Oct 23 '13 at 12:45
I'm sorry about being slow in understanding it- could you please explain how your final statement answers the problem? Thanks for your time! – Hummus Oct 23 '13 at 12:50
@Hummus, observe that we have started with $y=a^{\frac13}+b^{\frac13}$ and showed that $a^{\frac13}+b^{\frac13}$ satisfies $y^3=1-y$ – lab bhattacharjee Oct 23 '13 at 12:59
Thank you so much! Could you please explain how the final statement proves that y is a root of the cubic in the question? Thank you for your help! – Hummus Oct 23 '13 at 12:59
Right! I understand now! Thank you so much! – Hummus Oct 23 '13 at 13:03

robjohn · Answer 2 · 2013-10-23T13:22:47.100

Since $$ (x-a)(x-b)=x^2-(a+b)x+ab $$ we have $$ \color{#00A000}{a+b=1}\quad\text{and}\quad\color{#0000FF}{ab=-\frac1{27}} $$ therefore $$ \begin{align} \left(\color{#C00000}{a^{1/3}+b^{1/3}}\right)^3 &=a+3a^{2/3}b^{1/3}+3a^{1/3}b^{2/3}+b\\ &=\color{#00A000}{a+b}+3(\color{#0000FF}{ab})^{1/3}\left(\color{#C00000}{a^{1/3}+b^{1/3}}\right)\\ &=\color{#00A000}{1}+3\left(\color{#0000FF}{-\frac1{27}}\right)^{1/3}\left(\color{#C00000}{a^{1/3}+b^{1/3}}\right)\\ &=1-\left(\color{#C00000}{a^{1/3}+b^{1/3}}\right) \end{align} $$ Therefore, $$ \left(a^{1/3}+b^{1/3}\right)^3+\left(a^{1/3}+b^{1/3}\right)-1=0 $$

Roots of a Quadratic Problem

2 Answers2