use binomial theorem to expand (d-3b)^3?

Question

i need help with this problem, i dont understand any of the theorem or pascals triangle. please help. first answer gets best answer.

Answer 1

Ok, I'll bite...
You can get the coefficients of a binomial expansion from Pascal's Triangle, http://mathforum.org/dr.math/faq/faq.pascal.triangle.html
but that gets kind of awkward for higher powers, since you need to build Pascal's Triangle up to that level. For third power, tho, it works fine.
In general, the Binomial Theorem looks like this...
for (a + b)^n, the kth term is
C(n, (k-1)) a^(n-k-1) b^(k-1)
Let's break this down and see what it means...
k means the "kth" term. that is, the first term, second term, etc.
C(n, (k-1)) is the coeffients that you'd get from Pascal's Triangle. It can also be computed by factorials. Like this: http://mathforum.org/dr.math/faq/faq.comb.perm.html
(you probably have a combinations key or at least a factorials key on your calculator if you have a scientific or graphics calculator)
the a^(n-k-1) b^(k-1) mean the terms of the binomial expansion.
In other words, the first term of the binomial (in your example, d), starts at the higest power (in your example, 3) and goes down. While the second term (in your example, (-3b)) starts at 0 (in other words, it doesn't show up, since anything to the 0 is 1) and goes up to n (in your example, 3)
So, from Pascal's Triangle, the coeffiecnts are 1,3,3,1
To get these by the combination formula you'd do:
C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C(3,3) = 3
The terms would look like this:
1x(d^3)(-3b^0) + 3(d^2)(-3b^1) + 3(d^1)((-3b)^2) + 1(d^0)((-3b)^3)
this simplifies to
d^3 - 9d^2b + 27db^2 - 27b^3
(work this out and see if you can get to this simplification.)
Hope tihs helps...

Answer 2

Lonely has forgotten about one little detail, the minus signs. Those have to be dealt with when expanding the binomial. And sahsjing would have realized that -81b^3 can't possibly be the last term if he had simply cubed -3b in his head. It's equal to -27b^3, not -81b^3.

Combination theory is fine when one is dealing with binomial expansions to high powers, but for most problems you would find on an exam, simply memorizing Pascal's Triangle up to the 10th power is probably the most efficient route, because time is a limiting factor. It can be recreated quickly and easily on paper. See reference below. But don't limit yourself to just this. Seek out others too.

The coefficients of the terms of a binomial cubed are 1, 3, 3, 1, in that exact order.

We have here a situation where one of the terms is negative, so we use the regular format for a binomial cubed, then back substitute the values into it.

(x + y)^3 = x^3 + 3x^2 y + 3x y^2 + y^3.

Let x = d and y = -3b. Then the above expansion becomes:

d^3 + 3d^2 (-3b) + 3d (-3b)^2 + (-3b)^3 = d^3 - 9d^2 b + 27d b^2 - 27b^3.

So, the correct answer is:

(d - 3b)^3 = d^3 - 9d^2 b + 27d b^2 - 27b^3

Answer 3

Pascal's triangle can be used to find the coefficients in the binomial expansion. Numbering the top row as zero, and increasing downwards, that is the exponent.

For your question, we have the exponent 3, so we look at the fourth row (0, 1, 2, 3). That row is 1 3 3 1. Those coefficents get applied as we vary the powers of the variables inside the bracket.

So the answer to your question (the expansion), would be:

1*(d)^3*(3b)^0 + 3*(d)^2*(3b)^1 + 3*(d)^1*(3b)^2 + 1*(d)^0*(3b)^3

Simplifying gives:
d^3 + 3*d^2*(3b) + 3*d*(3b)^2 + (3b)^3

Answer 4

as per the binomial theorem
(x+y) ^n is given by x^n + nC1x^(n-1) y^1 + nC2x^(n-2) y^2 -------
Each term is given by T(r+1) = nCr x^(n-r) y^r
here C represents combination
like nC1 = n, nC2 = n(n-1)/2 and nC3 represent n(n-i)(n-2)/ (3)(2)(1)
In your question x = d, y= -3b , n= 3
Term 1 or T1 is found by putting r=0
T1 for your question is 3 C0 (d)^ (3-0) (-3b)^0 = d^3
T2 is given by 3 C1 (d)^ (3-1) (-3b)^1 = 3d^2 (-3b) = - 9 d^2 b
T3 is given by 3C2 (d)^ (3-2) (-3b)^2 = 3d^1 (-3b)^2 = 3 d^1 (9b^2)
27 d (b^2)
T4 is given by 3C3 (d)^ (3-3) (-3b)^3 = -27 b^3
(notice that the expansion has n+ 1 terms or 3+1 + 4 terms)
hence the expression
(d-3b) ^ 3 = [ d + (-3b)] ^ 3 = d^3 +(- 9 d^2)b+27 d (b^2) +(-27 b^3)
=d^3 - 9 d^2b+27 d (b^2) -27 b^3

Answer 5

Use Binomial Theorem To Expand

Answer 6

You can use combination concept to write binomial expansion.

(d-3b)^3
= (3C0)d^3 +(3C1)d^2(-3b) + (3C2)d(-3b)^2 + (3C3)(-3b)^3
= d^3 - 9d^2b + 27db^2 - 81b^3