Question 283682



{{{(3x+y)^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 {{{(3x+y)^3}}}, simply follow this procedure:

Write the first coefficient. Multiply that coefficient with the first binomial term {{{3x}}} and then the second binomial term {{{y}}}. Repeat this until all of the coefficients have been written.


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



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




However, we're not done yet.



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


{{{1(3x)^3(y)^0+(3x)^2(y)^1+3(3x)(y)+1(3x)(y)}}} Looking at the  second term {{{3(3x)(y)}}} raise  {{{3x}}} to the 2nd power and raise {{{y}}} to the 1st power.


Continue this until you reach the final term.



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



So the fully expanded expression should now look like this:



{{{1(3x)^3(y)^0+3(3x)^2(y)^1+3(3x)^1(y)^2+1(3x)^0(y)^3}}}



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



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



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



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



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