# Binomial theorem

six789
anyone could help me with this question...
in the expansion of (ax + by)^n, the coeffiients of the first 3 are 6561, 34992, and 81648., Find the value if a, b, and n.

i did this...
t1 = nC0 (ax)^n = 6561x^n
a^n = 6561

t2 = nC1 (ax)^n-1 (by)^1 = 34992x^n-1y
bna^n = 34992

but I'm not sure what to do next... and the rest

Homework Helper
bna^n = 34992
Double check this equation. It's close but it's wrong. The next thing to do should be obvious. They've given you the first 3 coefficients, and you've used 2 of them to find equations involving a, b, and n. As an extra hint, knowing (or rather assuming, but it seems implied) that a, b, and n are natural numbers, there are only four possibilities for the pair (a,n) given that an = 6561. You can find these possibilities if you find the prime factorization of 6561, it has a particularly nice one. One of these possibilities is (6561, 1), but you can eliminate this one immediately since it implies that n=1, but if this were the case, (ax + by)n would not even have 3 terms, but you're given the first 3 coefficients so it certainly has at least 3 terms. You're left with three possibilities. Sub one of them into your second equation and solve for b. When you find your third equation, see if you can sub in the a, b, and n that you now know in the left side, and see if it equals the right side. If it doesn't work, then eliminate that possibility, and you'll only have two left. Solve by process of elimination.

Homework Helper
Actually, you don't need to look at the prime factorization if you just do the algebra and solve the equations, I think.

EDIT: Yes, it's all just algebra. Make the right substitutions and solve the equations.

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Homework Helper
The prime factorization of all three coefficients makes it quite simple once you've got the three equations.

Homework Helper
Since you have three equations for the three unknown numbers, a, b, n, it isn't absolutely necessary that the numbers be integers, but it happens that the are and, as Shmoe says, prime factorization of the first number, at least, makes this problem very easy.

six789
guys, is it like this...?
for t2 = nC1 (ax)^n-1 (by)^1 = 34992x^n-1y
bna^n-1 = 34992

a^n = 6561
a^n = 3^8
therefore the answer for n = 8 and for a = 3, so you just have to find the b. but the second equation is bna^n-1 = 34992, so ill just substitute the values that i found...
it is like this...
bna^n-1 = 34992
b(8)(3)^8-1 = 34992
b = 2.

does my work lokks right now? anyone please check my mistake.

six789
guys, is it like this...?
for t2 = nC1 (ax)^n-1 (by)^1 = 34992x^n-1y
bna^n-1 = 34992

a^n = 6561
a^n = 3^8
therefore the answer for n = 8 and for a = 3, so you just have to find the b. but the second equation is bna^n-1 = 34992, so ill just substitute the values that i found...
it is like this...
bna^n-1 = 34992
b(8)(3)^8-1 = 34992
b = 2.

does my work looks right now? anyone please check my mistake.

Homework Helper
six789 said:
guys, is it like this...?
for t2 = nC1 (ax)^n-1 (by)^1 = 34992x^n-1y
bna^n-1 = 34992
This is the correct equation. You still need to produce a third equation using the third coefficient. Why haven't you?
a^n = 6561
a^n = 3^8
therefore the answer for n = 8 and for a = 3,
This is improper reasoning. Note that 38 = 94 = 81² = 65611 = 18.720754...³ so how do you know it's not one of these possibilities?

six789
is the thrid equation this...
b^2 n a^n-2 = 81,648

anyways, why do u need to find the third equation, since you have the two?

Homework Helper
six789 said:
is the thrid equation this...
b^2 n a^n-2 = 81,648

close, but check it again.

six789 said:
anyways, why do u need to find the third equation, since you have the two?

a^n=3^8 does not mean n=8 and a=3 (read AKG's last post again) so you'll need more information.

I should have mentioned in my last post that the prime factorizations make it very easy under the assumption a,b, n are integers. If you don't know this before hand, then some algebra is probably the best way to go.

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Homework Helper
As has been pointed out, the fact that an= 38 doesn't mean that a has to be 3 or that n has to be 8. However, if I were doing this problem my self, I would certainly notice and say "Aha, I wonder if that works!" You can use the second equation to find be assuming that a= 3 and n= 8. Then use the third equation to check to see if those values work.

Homework Helper
What Halls of Ivy says is right, but I don't think you can know a priori that just because one solution works that it is the only solution. That method will give you a solution, but will it tell you that there are no more? Again, if you solve it algebraically, you will find all solutions, and moreover you will not have to test any cases, or assume things are integers, etc.

six789
anyone guys, is the third equation b^2 2n a^n-2 = 81,648?

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Homework Helper
six789 said:
anyone guys, is the third equation b^2 2n a^n-2 = 81,648?
No, why '2n'?
$$\frac{n(n - 1)}{2!} a ^ {n - 2} b ^ 2 = 81648$$.
You should read the book again, to make sure that you understand everything. 