Hi,

Can you please help me with this question?

6(i) Expand (a+bx)3

The expansion of (1-x)(a+bx)3 in ascending powers of x begins 8-2x+…

(ii) Find the values of a and b.

(I found out that (a+bx)3 = a3 + 3a2bx + 3ab2x2 + b3x3 )

For (ii), do you have to substitute numbers into (1-x)(a+bx)3 ?

e.g.

sub x=0 into

(1-x) (a3 + 3a2bx) = 8-2x

a 3 = 8

a = 2

sub x=3 into (1-x) (23 + 3(2)2bx) = 8-2x

(1-3)(8+36b)=8-2(3)

-16-72b=2

72b=-18

b= -4??

Thank you so much!