Question

Expand the following using either the Binomial Theorem or Pascal’s Triangle. You must show your work for credit.

(2x + 4)^3

Answer 1

Binomial expansion works like this:
(a+b)^1 = (a+b)
(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2
(a+b)^3 = (a+b)(a2 + 2ab + b2) = a^3 + 3a^2 b + 3a b^2 + b^3

sub in your values for a and b from ur question above into:
a^3 + 3a^2 b + 3a b^2 + b^3and multiply out to get your answer

Answer 2

sorry, shorthand, that line should be:
(a+b)^3 = (a+b)(a+b)(a+b)
= (a+b)(a^2 + 2ab + b^2)
= a^3 + 3a^2 b + 3a b^2 + b^3

Answer 3

\[a^3 + 3a^2 b + 3a b^2 + b^3\]

Answer 4

\[\left( a+b \right)^{n}=nc0a ^{n}+nc1a ^{n-1}b+nc2a ^{n-2}b ^{2}+...+ncra ^{n-r}b ^{r}+...+ncnb ^{n}\]
\[nc0=1,nc1=\frac{ n }{1 }=n,nc2=\frac{ n \left( n-1 \right) }{ 2*1 },ncr=\frac{ n \left( n-1 \right)\left( n-2 \right)..\left\{ n-\left( r-1 \right) \right\} }{ r \left( r-1 \right)\left( r-2 \right)...3*2*1 }\]
\[ncn=\frac{ n \left( n-1 \right)\left( n-2 \right)...3*2*1 }{ n \left( n-1 \right) \left( n-2 \right)...3*2*1}=1\]
no. of terms=n+1
putn=3,a=2x,b=4 and get the solution.