How to expand the n-th power of (a + b) in brackets

a^n+C1/na^(n-1)b+C2/na^(n-2)b^2+…… +Cn-2/na^2b^(n-2)+Cn-1/nab^(n-1)+b^n
C1/n
1 on top
N is below

What is the negative power of the third square bracket of negative B?

Is this the case? (- B cubic) negative power? If this is the case, from the fact that the negative exponential power of any number that is not zero is the positive exponential power of its reciprocal, it can be known that it is 1 / 2 of - B cubic

9 (X-Y) 3 times square bracket - 1 / 9 (X-Y) m + 5 times square bracket (Y-X) 2 times
2 let m square + M = 1, find m3 + 2m square + 2005
Cubic power of M + 2m square + 2005
9 (X-Y) 3 power [- 1 / 9 (X-Y) m + 5 power] (Y-X) square
Square means square the letter in brackets
Why is 9 (X-Y) to the fifth power? Why is m square + m + 1 = 0?
It's M + 5

1
Power [- 5) y (x) - 9] + (x)
= 9 (X-Y) 3 power (Y-X) square) [- 1 / 9 (X-Y) m + 5 power]
=9 (X-Y) 5 power × [- 1 / 9 (X-Y) (M + 5) power]
=-1/(x-y)m
two
M square + M = 1
M square = 1-m
M3 power + 2m square + 2005
=M3 power + M2 + M2 + 2005
=M (m square + m) + m square + 2005
=m+1-m+2005
=2006

How to calculate the fourth power in brackets
For example, (I + 1) ^ 4 + (1-I) ^ 4

Because I ^ 2 = - 1
There is (I + 1) ^ 2 = I ^ 2 + 2I + 1 = - 1 + 2I + 1 = 2I
So (I + 1) ^ 4 = (2I) ^ 2 = - 4
And (1-I) ^ 2 = 1-2i-1 = - 2I
So (1-I) ^ 4 = (- 2I) ^ 2 = - 4
To sum up, (I + 1) ^ 4 + (1-I) ^ 4 = - 4 + (- 4) = - 8

How to expand the n-th power of (a + b) in brackets