Question 202859
Method # 1: The somewhat long way



{{{(x+3)^3}}} Start with the given expression.



{{{(x+3)(x+3)^2}}} Factor 



{{{(x+3)(x^2+6x+9)}}} FOIL



{{{x(x^2+6x+9)+3(x^2+6x+9)}}} Expand



{{{x^3+6x^2+9x+3x^2+18x+27}}} Distribute



{{{x^3+9x^2+27x+27}}} Combine like terms.




So {{{(x+3)^3}}} expands and simplifies to {{{x^3+9x^2+27x+27}}}.



In other words, {{{(x+3)^3=x^3+9x^2+27x+27}}}



<hr>


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



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




{{{(x+3)^3}}} Start with the given expression


To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle:
<center>1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;2&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;3&nbsp; &nbsp;3&nbsp; &nbsp;1&nbsp; &nbsp;</center>




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 {{{(x+3)^3}}}, 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:



{{{highlight(1)(x)(3)+highlight(3)(x)(3)+highlight(3)(x)(3)+highlight(1)(x)(3)}}} Notice how the coefficients are in front of each term.




However, we're not done yet.



{{{1(x)^3(3)^0+(x)(3)+3(x)(3)+1(x)(3)}}} Looking at the first term {{{1(x)(3)}}}, raise  {{{x}}} to the 3rd power and raise {{{3}}} to the 0th power.


{{{1(x)^3(3)^0+(x)^2(3)^1+3(x)(3)+1(x)(3)}}} Looking at the  second term {{{3(x)(3)}}} 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(x)^3(3)^0+3(x)^2(3)^1+3(x)^1(3)^2+1(x)^0(3)^3}}}



{{{1(x^3)(1)+3(x^2)(3)+3(x^1)(9)+1(x^0)(27)}}} Distribute the exponents



{{{1(x^3)+3(3x^2)+3(9x)+1(27)}}} Multiply



{{{x^3+9x^2+27x+27}}} Multiply the terms with their coefficients



So {{{(x+3)^3}}} expands and simplifies to {{{x^3+9x^2+27x+27}}}.



In other words, {{{(x+3)^3=x^3+9x^2+27x+27}}} (which is what we got before)