Binomial Expansion

Hi
Can you please help me with this problem?

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??

When I substituted a=2 and b=-4 into the equation it didn’t work.

Thank you

It is true that \(\displaystyle a = 2\). To find \(\displaystyle b\), you must solve \(\displaystyle 3a^2b - a^3 = -2\) for \(\displaystyle b\), when \(\displaystyle a = 2\). Be sure you understand why.