00:01
In this question, we have an equation x to the power 11 minus 1 is equal to 0.
00:11
Alpha is equal to cos 2 pi by 11 plus iota sign 2 pi by 11 and beta is equals to alpha cube.
00:28
We need to find how many of the given options are correct.
00:34
Since the coefficients of the equation are real complex roots if any occur in conjugate pairs, therefore the roots of the equation can be written as roots are 1, cos 2k pi by 11 plus minus iota sign 2k pi by 11, where k is equals to 1, 2, 3, 4, 4.
01:13
And 5.
01:18
Now since cos 2k pi by 11 minus iota sign 2k pi by 11 is equals to 1 upon cos 2k pi by 11 plus iota 2k pi by 11.
01:51
Therefore, roots can be represented as represented as 1, alpha, alpha square, alpha cube, alpha to the power 4, alpha to the power 5, 1 upon alpha, 1 upon alpha and 1 upon alpha to the power 5.
02:35
Thus option a is correct.
02:43
Option a is correct.
02:51
Now let's check the option b.
03:00
Since beta is equal to alpha cube, therefore we can say that beta cube is equal to alpha to alpha to the power 9 and this will be equals to alpha to the power 11 upon alpha square.
03:24
And this will be equal to 1 upon alpha square...