Find the value of $\alpha^{\frac13}+\beta^{\frac13}$

Question

If $f(x)=x^2-5x+8, f(\alpha)=0$ and $f(\beta)=0$ then find the value of $\alpha^{\frac13}+\beta^{\frac13}$

$$\alpha+\beta=5$$ $$\alpha \beta=8$$ $$\alpha^{\frac 1 3}=\frac 2 {\beta^{\frac 1 3}}$$

what should I do next?

Answer 1

Let $(\alpha)^{1/3}=c,(\beta)^{1/3}=d$ then $(c+d)^3=c^3+d^3+3cd(c+d)$ thus we have $(c+d)^3=k_1+k_2(c+d)$ where $k_1=\alpha+\beta,k_2=3((\alpha)(\beta))^{1/3}$ let $c+d=y$ then we have $y^3=5+6y$ we can see that $-1$ is one of the solutions for y other two can be found out by polynomial division.

Answer 2

Since $f(\alpha)=0$ and $f(\beta)=0$, so $\alpha, \beta$ are solutions of the given equation $f(x)=x^2-5x+8=0$.

So we say that $\alpha=\frac{5 + i \sqrt{7}}{2}$, $\beta=\frac{5 - i \sqrt{7}}{2}$.

Also $\alpha + \beta= 5$, $\alpha \beta =8$

Let $\alpha^{\frac{1}{3}}=a$ and $\beta^{\frac{1}{3}}=b$

Now $(a+b)^3 =a^3 + b^3+3ab(a+b)\implies (\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})^3= \alpha + \beta + 3\alpha^{\frac{1}{3}}\beta^{\frac{1}{3}}(\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}}) \implies (\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})^3 =5+6(\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})\implies m^3-6m-5=0$ (taking $(\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})=m$)

Now $m^3-6m-5=0 \implies (m+1)(m^2-m-5)=0\implies m=-1, \frac{1\pm \sqrt{21}}{2}$

Hence values of $\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}}$ are $-1, \frac{1\pm \sqrt{21}}{2}$.