Difficulty in finding the roots of a cubic equation.

Question

Find the roots $\alpha,\beta, \gamma$ of $x^3 -11x^2+36x-36=0$ if $\dfrac{2}{\beta}= \dfrac{1}{\alpha}+\dfrac{1}{\gamma}$

Now, I got the following equations using sum of roots and product of roots "formulae" for a polynomial with n roots:

$\alpha+\beta+\gamma=11$

$\alpha\beta\gamma=36$

There are 3 equations and three variables but I am facing difficulty in solving these three equations. Like, I managed to get following two equations:

$\alpha^2\gamma+\alpha\gamma^2+36-11\alpha\gamma=0$

$\alpha^2+4\alpha\gamma+\gamma^2+11\gamma-11\alpha=0$

But these two are threatening equations. How do I continue or is there an easier approach to this problem?

Answer 1

Hint: $$ \frac{3}{\beta} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} $$ and you know $\alpha\beta + \beta\gamma + \gamma\alpha = 36$ for the same reason you know $\alpha + \beta + \gamma$ or $\alpha \beta \gamma$

Answer 2

HINT: by the rational root Theorem we have $$x_1=2,x_2=3,x_3=6$$

Answer 3

$\alpha \beta \gamma = 36$

$\alpha + \beta + \gamma = 11$

$ \alpha \beta+ \beta \gamma + \alpha \gamma =36$

$ (\alpha + \gamma) \beta = 2 \alpha \gamma$

If you manipulate these four equalitites, you get

$\alpha =6$ $\beta = 3$ $\gamma = 2$

Answer 4

Hint:

Let $\alpha=\dfrac1a$ etc.

$\implies2b=a+c$

So, $a,b,c$ are the roots of $$36y^3-36y^2+11y-1=0$$

$\implies\dfrac{36}{36}=a+b+c=3b\iff b=\dfrac13$

Divide $36y^3-36y^2+11y-1$ by $3y-1$ to find a quadratic equation.