Question

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

• A. $-6-6{\omega }^2$
• B. $1+6\omega +12{\omega }^2$
• C. $-5+6{\omega }^2$
• D. None of these

Answer 1

Answer: (B), (C)

$-5+6{\omega }^2$
$1+6\omega +12{\omega }^2$

$x^3-1=0$

non real roots of the above equation are $\alpha =w\&\beta =w^2$.

where $w$ is the cube root of unity

$\Rightarrow w^3=1\&1+w+w^2=0$
$z={(1+2\alpha +3\beta )}^3{-(3+3\alpha +5\beta )}^3$

$\Rightarrow z={(1+2w+3w^2)}^3-(3+3w+5w^2)$

$\Rightarrow z={(-w-w^2+2w+3w^2)}^3-{(-3w^2+5w^2)}^3$

$\Rightarrow z={(w+{2w}^2)}^3-{\left(2w^2\right)}^3$

$\Rightarrow z=w^3+8w^6+6w^4+12w^5-8w^6$

$\Rightarrow z=1+6w+12w^2$

$\Rightarrow z=-5+6w^2$