Answer :
[tex](c-2d)^5=(-32)d^5+80cd^4-80c^2d^3+40c^3d^2-10c^4d+c^5\\\\the \ common \ formula\ is:\\(a-b)^2=-b^5+5ab^4-10a^2b^3+10a^3b^2-5a^4b+a^5[/tex]
Binomials can be expanded using numbers in Pascal's triangle, with the row in the triangle being determined by the power, in this case 5.
Each term in the expansion of (x+y)^n is represented byax^by^c
With a being the number in that position of Pascal's triangle,
b being n - the position in the triangle
c being the position in the triangle,
And x and y being the terms used in the original
(x+y)^3 uses the row 1 3 3 1
So
1x^3y^0+3x^2y^1+3x^1y^2+1x^0y^3
Or x^3+3x^2y+3xy^2+y^3
Answering your question:
In your case, we will use the fifth row, which is 1 5 10 10 5 1
And we get
1c^5(-2d)^0+5c^4(-2d)^1+10c^3(-2d)^2+10c^2(-2d)^3+5c^1(-2d)^4+1c^0(-2d)^5
This can be simplified to end up with
C^5-10c^4d^5+40c^3d^2-80c^2d^3+80cd^4-32d^5
Which is the final answer
Hope this helps :)
Each term in the expansion of (x+y)^n is represented byax^by^c
With a being the number in that position of Pascal's triangle,
b being n - the position in the triangle
c being the position in the triangle,
And x and y being the terms used in the original
(x+y)^3 uses the row 1 3 3 1
So
1x^3y^0+3x^2y^1+3x^1y^2+1x^0y^3
Or x^3+3x^2y+3xy^2+y^3
Answering your question:
In your case, we will use the fifth row, which is 1 5 10 10 5 1
And we get
1c^5(-2d)^0+5c^4(-2d)^1+10c^3(-2d)^2+10c^2(-2d)^3+5c^1(-2d)^4+1c^0(-2d)^5
This can be simplified to end up with
C^5-10c^4d^5+40c^3d^2-80c^2d^3+80cd^4-32d^5
Which is the final answer
Hope this helps :)
