If alpha, beta and gamma are the roots of the equation x^3+6x+1=0, how do you find the polynomial whose roots are alpha beta , beta gamma and alpha gamma?

Sep 24, 2017

The equation is ${x}^{3} - 6 {x}^{2} - 1 = 0$

Explanation:

Let's rewrite the equation as

${x}^{3} + 0 {x}^{2} + 6 {x}^{1} + 1 \cdot {x}^{0} = 0$

If $\alpha$, $\beta$ and $\gamma$ are the roots of this equation, we have

$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) = 0$

$\left({x}^{2} - \alpha x - \beta x + \alpha \beta\right) \left(x - \gamma\right) = 0$

${x}^{3} - \left(\alpha + \beta + \gamma\right) {x}^{2} + \left(\alpha \beta + \gamma \beta + \alpha \gamma\right) x - \alpha \beta \gamma = 0$

Comparing this equation to the original equation,

$\alpha + \beta + \gamma = 0$

$\alpha \beta + \gamma \beta + \alpha \gamma = 6$

$\alpha \beta \gamma = - 1$

The new roots are,

$\alpha \beta$, $\beta \gamma$ and $\alpha \gamma$

Therefore,

$\alpha \beta + \beta \gamma + \alpha \gamma = 6$

The coefficient of ${x}^{2}$ is $- 6$

$\alpha \beta \cdot \beta \gamma \cdot \alpha \gamma = {\left(\alpha \beta \gamma\right)}^{2} = {\left(- 1\right)}^{2} = 1$

The coefficient of ${x}^{0}$ is $- 1$

$\alpha \beta \cdot \beta \gamma + \alpha \gamma \cdot \alpha \beta + \alpha \beta \cdot \beta \gamma = \alpha \beta \gamma \left(\alpha + \beta + \gamma\right) = 0$

The coefficient of ${x}^{1}$ is $0$

The equation is

${x}^{3} - 6 {x}^{2} + 0 {x}^{1} - 1 \cdot {x}^{0} = 0$

$\implies$, ${x}^{3} - 6 {x}^{2} - 1 = 0$

Sep 24, 2017

${x}^{3} - 6 {x}^{2} - 1 = 0.$

Explanation:

We know that, $\alpha , \beta , \gamma$ are the roots of the eqn.

${x}^{3} + 0 {x}^{2} + 6 x + 1 = 0.$

Then, by Vieta's Rule, we have,

(1):alpha+beta+gamma=0; (2):alphabeta+betagamma+gammaalpha=6;(3):alphabetagamma=-1.

Let, $\alpha \beta = c , \beta \gamma = a , \gamma \alpha = b .$

Then, $a , b , c$ are the roots of the desired poly.

Now, $\left(1 '\right) : a + b + c = \beta \gamma + \gamma \alpha + \alpha \beta = 6 ,$

$\left(2 '\right) : a b + b c + c a = \alpha \beta {\gamma}^{2} + {\alpha}^{2} \beta \gamma + \alpha {\beta}^{2} \gamma ,$

$= \alpha \beta \gamma \left(\alpha + \beta + \gamma\right) = - 1 \cdot 0 = 0 ,$ and,

$\left(3 '\right) : a b c = \left(\beta \gamma\right) \left(\gamma \alpha\right) \left(\alpha \beta\right) = {\alpha}^{2} {\beta}^{2} {\gamma}^{2} = {\left(- 1\right)}^{2} = 1.$

From $\left(1 '\right) , \left(2 '\right) , \mathmr{and} , \left(3 '\right) ,$ the desired poly cubic eqn. is,

${x}^{3} - \left(a + b + c\right) {x}^{2} + \left(a b + b c + c a\right) x - a b c = 0 , i . e . ,$

${x}^{3} - 6 {x}^{2} + 0 x - 1 = 0 , \mathmr{and} , {x}^{3} - 6 {x}^{2} - 1 = 0.$

Enjoy Maths.!