Question

If 1, $\omega$ , ${\omega }^2$ are the roots of the $x^3$ + p$x^2$ + qx + r = 0 , find ( p , q, r)

• A. (1, –1, 0)
• B. (1, 1, –1)
• C. (0, 0, –1)
• D. (0, –1, 0)

Answer 1

The correct option is C

(0, 0, –1)

1, $\omega$ , ${\omega }^2$ are the cube roots of unity.
1, $\omega$ , ${\omega }^2$ are the roots of
$x^3$ = 1
OR
$x^3$ - 1 = 0
Comparing it with the equation in question, we get
p = 0 , q = 0 , r = -1
Alternate solution :
We know 1 + $\omega$ + ${\omega }^2$ = 0
As 1, $\omega$ , ${\omega }^2$ are roots of $x^3$ + p$x^2$ + qx + r = 0
- p = Sum of roots = 1 + $\omega$ + ${\omega }^2$ = 0
Option A and B are eliminated
Also we know ${\omega }^3$ = 1
-r = 1 r = -1
So, (c) is right option