Question

If $x^3-1=0$ has the non real complex roots a, b then the value of $(1+2\alpha +\beta {)}^3-(3+3\alpha +5\beta {)}^3$ is?

• A. $-7$
• B. $6$
• C. $-5$
• D. $0$

Answer 1

The correct option is A $-7$

here given $x^3-1=0$

has roots $\alpha$ & $\beta$ which are complex.

x = $\alpha$ & $\alpha =\omega$

x = $\beta$ & $\beta ={\omega }^2$

now, ${(1+2\alpha +\beta )}^3-{(3+3\alpha +5\beta )}^3$

= ${(1+2\omega +{\omega }^2)}^3-{(3+3\omega +5{\omega }^2)}^3$

= ${(1+\omega +{\omega }^2+\omega )}^3-{(3(1+\omega +{\omega }^2)+2{\omega }^2)}^3$

as we know, $1+\omega +{\omega }^2=0$

= ${(0+\omega )}^3-{(3\left(0\right)+2{\omega }^2)}^3$

= ${\omega }^3-8{\omega }^6$

= 1 - 8 $({\omega }^3=1)$

= -7

so, ${(1+2\alpha +\beta )}^3-{(3+3\alpha +5\beta )}^3=-7$