Topic: Binomial Theorem (finding ratio of unknowns) [unsolved]

[Auxuris]

In the expansion of (1+x)(a-bx)¹², where ab≠0, the coefficient of x⁸ is zero. Find, in its simplest form, the value of ratio a/b.

So I tried comparing coefficients, and I got that I'd need to expand an x⁷ and x⁸ to get the total coefficient of x⁸ in order to equate that to zero.


But the fractions just didn't work out.
Does anyone have a more efficient method?
Or if you could would this/any method out accurately because I can't seem to do it.

ps i'm revising this topic when i came across this question

 

[HallsofIvy]

In the expansion of (1+x)(a-bx)¹², where ab≠0, the coefficient of x⁸ is zero. Find, in its simplest form, the value of ratio a/b.

So I tried comparing coefficients, and I got that I'd need to expand an x⁷ and x⁸ to get the total coefficient of x⁸ in order to equate that to zero.


But the fractions just didn't work out.
Does anyone have a more efficient method?
Or if you could would this/any method out accurately because I can't seem to do it.

ps i'm revising this topic when i came across this question

That method certainly does work. Please show your work. What did you get for the coefficient of x⁷ in (a- bx)¹²? What did you get for the coefficient of x⁸?