Question

WILL GIVE MEDAL use binomial theorem to expand the binomial (d-4b)^3

Answer 1

@hartnn help pls lol

Answer 2

state the binomial thrm ... its assuming you have some prior knowledge for the question

Answer 3

i dont understand it though, i understand the Pascal Tri, but not this

Answer 4

the pascal triangle is just a visual representation of the binomial thrm ... a set of coefficients

Answer 5

\[(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}~b^{k}\]
this is a compact notation for a sum of a lot of terms

Answer 6

okay im lost

Answer 7

if we use the expanded version of it ...

\[(a+b)^n=\binom{n}{0}a^{n}~b^{0}+\binom{n}{1}a^{n-1}~b^{1}+\binom{n}{2}a^{n-2}~b^{2}+...+\binom{n}{n}a^{n-n}~b^{n}\]

Answer 8

what are you lost on?

Answer 9

everything

Answer 10

can you at least fill in the parts? they give you n,a, and b parts

Answer 11

ok

Answer 12

-64b^3+48b^2d-12bd^2+d^3

Answer 13

would that be it?

Answer 14

maybe .. do you have to do this by the binomial setup? or do they just expect a final result?

Answer 15

i have to do the binomial setup

Answer 16

ik the answer, but i want to know how to do it

Answer 17

@Pleasehelpasap12345 will help you in a second, msgs wont work

Answer 18

ok

Answer 19

let a = d
let b = -4b
let n = 3
\[(a+b)^n=\binom{n}{0}a^{n}~b^{0}+\binom{n}{1}a^{n-1}~b^{1}+\binom{n}{2}a^{n-
2}~b^{2}+...+\binom{n}{n}a^{n-n}~b^{n}\]

\[(d+(-4b))^3=\binom{3}{0}d^{3}~(-4b)^{0}+\binom{3}{1}d^{3-1}~(-4b)^1\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+\binom{3}{2}d^{3-2}~(-4b)^{2}+\binom{3}{3}d^{3-3}~(-4b)^{3}\]

\[(d+(-4b))^3=\binom{3}{0}d^{3}-\binom{3}{1}4d^{2}b+\binom{3}{2}16db^{2}-\binom{3}{3}64b^{3}\]
the parantheticals just relate to the pascal row values
\[(d+(-4b))^3=(1)d^{3}-(3)4d^{2}b+(3)16db^{2}-(1)64b^{3}\]

Answer 20

ok

Answer 21

still pretty confusing

Answer 22

thats all it is ... filling in the parts good luckl