00:01
Let's write the partial fraction decomposition of 2 over xq minus 1.
00:06
So first i'm going to write down the denominator expression, that is xq minus 1.
00:12
Using identity, we write this as x minus 1 times x squared plus x plus 1.
00:22
We basically used this identity, that is a cube minus b cube.
00:28
This equals a minus b, times a squared.
00:33
Plus ab plus b squared.
00:38
So when we use this identity, we can rewrite this xq minus 1 as this one.
00:44
Now we write this as a sum of partial fractions.
00:47
We're seeing partial fraction decomposition methods.
00:50
So we can write this as a over x minus 1.
00:54
And since this term cannot be factored, we write this as bx plus c, toded by x squared plus x plus 1.
01:04
Now we should focus on this equation so that we can solve for a, b, and c.
01:12
So let's multiply both sides by the common denominator, which is this expression.
01:18
And so when you do that on the left, we will be left with the two.
01:21
And this equals a will be multiplied with the term, which it doesn't have in the denominator compared with the common denominator.
01:31
So it does not have this expression that is x squared plus x plus 1.
01:36
So this will be multiplied with x squared plus x plus 1.
01:41
And then the next term bx plus c, this will be multiplied with the term x minus 1.
01:50
Let's simplify this and solve for a b and c.
01:54
So here we have 2.
01:55
This equals let's distribute this.
01:58
This will be a x squared plus a x x.
02:03
Plus a, then we will also follow this.
02:09
So bx times x is bx squared and bx times negative 1 is negative bx.
02:18
Then c times x is positive c x and c times negative 1 is negative c.
02:26
And then next step we are going to group the light terms.
02:31
We have x squared terms.
02:33
So when you group them and factor x squared we will get a plus b times x squared now let's group the x term here we have a x negative bx plus cx and so we factor x which will be a this will be negative b a negative b and then now pause it c times x scoop the constant term that is a minus c so the constant term is a minus c.
03:06
Now we can equate the coefficients on the left side we don't have x squared so it's equivalent to 0 x squared as plus 0 plus 2 which means a plus p equals 0 and we get another equation a minus p plus c this is also 0 because the coefficient of x is also 0 over here and finally a minus c this equals 2 so now we have are three equations which is a system of equations in the variables a b and c...