SOLUTION: i was wondering how would you do this problem and find its products : (x+3)^3 and thank you in advance for helping me

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Question 202859: i was wondering how would you do this problem and find its products :
(x+3)^3
and thank you in advance for helping me
Found 2 solutions by jim_thompson5910, RAY100:
Answer by jim_thompson5910(28598) About Me  (Show Source):
You can put this solution on YOUR website!
Method # 1: The somewhat long way


%28x%2B3%29%5E3 Start with the given expression.


%28x%2B3%29%28x%2B3%29%5E2 Factor


%28x%2B3%29%28x%5E2%2B6x%2B9%29 FOIL


x%28x%5E2%2B6x%2B9%29%2B3%28x%5E2%2B6x%2B9%29 Expand


x%5E3%2B6x%5E2%2B9x%2B3x%5E2%2B18x%2B27 Distribute


x%5E3%2B9x%5E2%2B27x%2B27 Combine like terms.



So %28x%2B3%29%5E3 expands and simplifies to x%5E3%2B9x%5E2%2B27x%2B27.


In other words, %28x%2B3%29%5E3=x%5E3%2B9x%5E2%2B27x%2B27




If you didn't like method # 1, then...


Method # 2: The shorter way (if you're familiar with this method)



%28x%2B3%29%5E3 Start with the given expression

To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle:
1   

1   1   

1   2   1   

1   3   3   1   




Looking at the row that starts with 1,3, etc, we can see that this row has the numbers:

1, 3, 3, and 1

These numbers will be the coefficients of our expansion. So to expand %28x%2B3%29%5E3, simply follow this procedure:
Write the first coefficient. Multiply that coefficient with the first binomial term x and then the second binomial term 3. Repeat this until all of the coefficients have been written.

Once that has been done, add up the terms like this:


Notice how the coefficients are in front of each term.



However, we're not done yet.


1%28x%29%5E3%283%29%5E0%2B%28x%29%283%29%2B3%28x%29%283%29%2B1%28x%29%283%29 Looking at the first term 1%28x%29%283%29, raise x to the 3rd power and raise 3 to the 0th power.

1%28x%29%5E3%283%29%5E0%2B%28x%29%5E2%283%29%5E1%2B3%28x%29%283%29%2B1%28x%29%283%29 Looking at the second term 3%28x%29%283%29 raise x to the 2nd power and raise 3 to the 1st power.

Continue this until you reach the final term.


Notice how the exponents of x are stepping down and the exponents of 3 are stepping up.


So the fully expanded expression should now look like this:





1%28x%5E3%29%281%29%2B3%28x%5E2%29%283%29%2B3%28x%5E1%29%289%29%2B1%28x%5E0%29%2827%29 Distribute the exponents


1%28x%5E3%29%2B3%283x%5E2%29%2B3%289x%29%2B1%2827%29 Multiply


x%5E3%2B9x%5E2%2B27x%2B27 Multiply the terms with their coefficients


So %28x%2B3%29%5E3 expands and simplifies to x%5E3%2B9x%5E2%2B27x%2B27.


In other words, %28x%2B3%29%5E3=x%5E3%2B9x%5E2%2B27x%2B27 (which is what we got before)

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
(x+3)^3 = (x+3)(x+3)(x+3)
.
We can foil one set,,but then distribute last
.
(x+3) (x^2 +6x+9)
.
(x^3 +6x^2 +9x)+ (3x^2 +18x +27)
.
x^3 +9x^2 +27x +27
,
check ,,,let x=1,,,64 = original and answer,,,ok