Expand: (x+3) to the power of 3?

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Expand: (x+3) to the power of 3?

Mathematics
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1331
x^3 +3x^2(3) +3x(3^2) + 3^3
x^3 +9x^2 +27x +27

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Start off on the basic. (x+3)(x+3) = x^2 + 6x + 9 Then multiply through again. (x^2 + 6x + 9)(x + 3) x^3 + 6x^2 + 9x + 3x^2 + 18x + 27 x^3 + 9x^2 + 27x + 27
i think its part of the binomial theorum; which has the coefficients of Pascals triangle. At ^3 we get coefs of 1,3,3,1
then just input the first and last terms last term decreases by ^1 and first term increases by^1
(x+3)^5 1 5 10 10 5 1 1x^5 +5x^4 +10x^3 +10x^2 +5x +1 1x^5 +5x^4(3) +10x^3(x^2) +10x^2(3^3) +5x(3^4) +1(3^5)
x^5 +15x^4 +90x^3 +270x^2 +405x +3^5 lol
They'll come a point when the power is too high for you to continue your precious triangle. Or too non-positive-integery
yeah, but its fun while it lasts :) my son thinks its cool lol
i'm lost... do you basically foil 3 times or what?
yes, "FOIL" 3 times, but that last "FOIL" becomes more of a "FOIOIOL" or some such nonesense...
distribute 3 times is a better acounting for it
I tell my kids, no you dont "FOIL" anything. You work it out thru the distributive property. And once you know what is going on....name it whatever you want :)
what would the final answer come out to ?
its already been posted..... scroll up :)
or... distribute it out 3 times....
tell me the answer again... i dont know which one the answer is?? pleaz write it again

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