# If alpha=(1/2)(-1+sqrt(-3)) and beta=(1/2)(-1-sqrt(-3)) then prove alpha^4+(alphabeta)^2+beta^4=0?

Aug 12, 2018

#### Explanation:

Here ,

$\alpha = \frac{1}{2} \left(- 1 + \sqrt{- 3}\right) \mathmr{and} \beta = \frac{1}{2} \left(- 1 - \sqrt{- 3}\right)$

$\therefore \alpha + \beta = \frac{1}{2} \left\{- 1 + \sqrt{- 3} - 1 - \sqrt{- 3}\right\} = \frac{1}{2} \left(- 2\right) = - 1$

$\alpha \cdot \beta = \frac{1}{4} \left\{{\left(- 1\right)}^{2} - {\left(\sqrt{- 3}\right)}^{2}\right\} = \frac{1}{4} \left\{1 - \left(- 3\right)\right\}$=$\frac{1}{4} \left\{4\right\} = 1$

i.e. color(red)(alpha+beta=-1 and alpha*beta=1to(1)

We know that ,

${\left(\alpha + \beta\right)}^{2} = {\alpha}^{2} + {\beta}^{2} + 2 \alpha \beta$

$\therefore {\left(- 1\right)}^{2} = {\alpha}^{2} + {\beta}^{2} + 2 \left(1\right) \to F r o m \left(1\right)$

:.color(blue)(alpha^2+beta^2=1-2=-1to(2)

We also know that ,

${\left({\alpha}^{2} + {\beta}^{2}\right)}^{2} = {\alpha}^{4} + {\beta}^{4} + 2 {\alpha}^{2} {\beta}^{2}$

$\therefore {\left(- 1\right)}^{2} = {\alpha}^{4} + {\beta}^{4} + {\left(\alpha \beta\right)}^{2} + {\left(\alpha \beta\right)}^{2} \to F r o m \left(2\right)$

$\therefore {\alpha}^{4} + {\beta}^{4} + {\left(\alpha \beta\right)}^{2} = {\left(\alpha \beta\right)}^{2} - 1$

$\therefore {\alpha}^{4} + {\beta}^{4} + {\left(\alpha \beta\right)}^{2} = {\left(1\right)}^{2} - 1 = 0$