If $2\alpha = - 1 - i\sqrt{3}$ and $2\beta = - 1 + i\sqrt{3}$, then $5\alpha^{4} + 5\beta^{4} + 7\alpha^{- 1}\beta^{- 1}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Given, $2\alpha = - 1 - i\sqrt{3}$ and $2\beta = - 1 + i\ \sqrt{3}$
$	herefore\ \ \alpha + \beta = - 1$ and $\alpha\beta = 1$
Now, $5\alpha^{4} + 5\beta^{4} + \frac{7}{\alpha\beta}$
$= 5\lbrack\left\{ (a + \beta)^{2} - 2\alpha\beta ight\}^{2} - (\alpha\beta)^{2} + \frac{7}{\alpha\beta}$
$= 5\lbrack\left\{ ( - 1)^{2} - 2 	imes 1 ight\}^{2} - 2(1)^{2}brack + \frac{7}{1}$
$= 5(1 - 2) + 7 = 2$

Q. If $2\alpha = - 1 - i\sqrt{3}$ and $2\beta = - 1 + i\sqrt{3}$, then $5\alpha^{4} + 5\beta^{4} + 7\alpha^{- 1}\beta^{- 1}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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