# Binomial expansion- Constant term(Question 2

You could (roughly) write a few of the terms down and see if you can spot which term it corresponds to. You could pull the 1/x^2 outside which might simplify it slightly.
(edited 7 months ago)

Go for the freebies first (despite being a multiple choice and no marks for workings).
In this case, the freebie is "can you write down the binomial theorem for this particular problem".
The next one, slightly less freebie but still crucial, is "using above, can you locate the constant term".

Also I don't know what you mean by "no single digit of x", but probably not a very useful line.

post what you have tried

(a-b)^12 = sum from n=0 to 12 (-1)^n(12Cn)a^(12-n)b^n
so we want the constant term, which is x^0
so what a^(12-n)b^n will give that?
a=x^6,b=x^-2

Spoiler

(edited 7 months ago)
Original post by mqb2766
You could (roughly) write a few of the terms down and see if you can spot which term it corresponds to. You could pull the 1/x^2 outside which might simplify it slightly.

when i tried it i said 1/x^2= x^-2 but then i was like how on earth can this help me if i don't have a constant because you can't do binomial expansion without iy
Original post by realed
when i tried it i said 1/x^2= x^-2 but then i was like how on earth can this help me if i don't have a constant because you can't do binomial expansion without iy

If you have
(a+b)^n
then the expansion is
a^n + nC1*a^(n-1)b + nC2*a^(n-2)b^2 + .... + b^n
So when you multipy by "b"s the power of x gets reduced. Iif you wrote out a few of the terms you should see this. The question is asking for the coeffiicent of the term x^0.

Or alternatively factor out 1/x^(2) so
1/x^(24) (x^8 - 1)^12
and for the constant term youre looking for the term in (x^8 - 1)^12 which corresponds to x^24.
(edited 7 months ago)
if i factor x^-2 out i get (x^-2(X^8-1))^12

but what on earth happens next
Original post by realed
if i factor x^-2 out i get (x^-2(X^8-1))^12

but what on earth happens next

You factor x^(-2) out of each of the 12 terms. So the overall multiplier is x^(-24).
x^-24 * (X^8-1)^12

Then you pick the term corresponding to x^(24) in (X^8-1)^12, as when you multiply by x^(-24) it will give the constant term or the x^0 term.
(edited 7 months ago)
Original post by mqb2766
You factor x^(-2) out of each of the 12 terms. So the overall multiplier is x^(-24).
x^-24 * (X^8-1)^12

Then you pick the term corresponding to x^(24) in (X^8-1)^12, as when you multiply by x^(-24) it will give the constant term or the x^0 term.

wait so would that be (x^8-1)^3=-220 which is the answer as x^24 and x^-24 cancel out
Original post by realed
wait so would that be (x^8-1)^3=-220 which is the answer as x^24 and x^-24 cancel out

Not sure where the 3 in (x^8-1)^3 comes from, but the answer is correct. Youre looking for the term corresponding to (x^8)^3 in (x^8-1)^12. Expanding the latter gives the 13 terms in the usual binomial expansion
x^96 - 12C1 x^88 + 12C2 x^80 - .... + 12C10 x^16 - 12C11 x^8 + 1
So its the 10th term (corresponding to x^24) which has coefficient -12C9, which is -220 as you say.

Its not an unusual binomial expansion question so if youre hitting tmua in oct, it may be worth reviewing the basic binomial stuff as its reasonably common to factor out terms like 1/x^2 and be able to write down the terms of the remaining binomial (x^8 - 1)^12 to identify which part you want. Similarly being aware of the expansion of (a+b)^n is slightly more direct.
(edited 7 months ago)