Question

If $(z^3$ + $\frac{1}{z^3})$ - $(z^2$ + $\frac{1}{z^2})$ + $(z+\frac{1}{z})$ - 1 = $(z+\frac{1}{z}-2\alpha$) $(z+\frac{1}{z}-2\beta$) $(z+\frac{1}{z}-2\gamma$) Find the value of $\alpha$.$\beta$.$\gamma$.

• A. .
• B. .
• C. .
• D. .

Answer 1

The correct option is C

.

$(z^3$ + $\frac{1}{z^3})$ - $(z^2$ + $\frac{1}{z^2})$ + $(z+\frac{1}{z})$ - 1 = $(z+\frac{1}{z}-2\alpha$) $(z+\frac{1}{z}-2\beta$) $(z+\frac{1}{z}-2\gamma$)---------(1)

Lets assume $z+\frac{1}{z}$ = 2x--------(2)

$(z+\frac{1}{z}{)}^2$ = $z^2$ + $\frac{1}{z^2}$ + 2($z\times \frac{1}{z})$

$z^2$ + $\frac{1}{z^2}$ = $(2x{)}^2$ - 2 = 4$x^2$ - 2 ------------(3)

$(z+\frac{1}{z}{)}^3$ = $z^3$ + $\frac{1}{z^3}$ + 3($z+\frac{1}{z})$

$z^3$ + $\frac{1}{z^3}$ = $(2x{)}^3$ - 3 $\times \left(2x\right)$ = 8$x^3$ - 6x-----------(4)

Substitute the value of $(z+\frac{1}{z})$, $z^2$ + $\frac{1}{z^2}$ and $z^3$ + $\frac{1}{z^3}$ in equation (1)

$(8x^3-6x)-(4x^2-2)+\left(2x\right)-1=(2x-2\alpha$) (2x - 2$\beta$) (2x - 2$\gamma$)

$8x^3$ - $4x^2$ - 4x + 1 = 8(x - $\alpha$) (x - $\beta$) (x - $\gamma$)--------------(5)

From equation 5,we see that $\alpha$,$\beta$ and $\gamma$ are the roots of the equation $8x^3$ - $4x^2$ - 4x + 1 = 0

Product of the roots $\alpha$,$\beta$,$\gamma$ = $\frac{-1}{8}$.