Question 1091393
<br>This is a type of algebra problem commonly encountered on high school math contests.<br>
You need to use the given equations somehow to get an expression containing x^3 and y^3.  One way to do this is to cube the first equation:<br>
{{{(x+y)^3 = x^3 + 3(x^2)y + 3xy^2 + y^3}}}
{{{(x+y)^3 = (x^3 + y^3) + 3xy(x+y)}}}
{{{x^3 + y^3 = (x+y)^3 - 3xy(x+y)}}}<br>
You know the value of (x+y); to get a numerical value for this expression, you need to find the value of xy.  To do that, you can square the first equation:<br>
{{{(x+y)^2 = x^2 + 2xy + y^2 = (x^2 + y^2) + 2xy = 9}}}
{{{6 + 2xy = 9}}}
{{{2xy = 3}}}
{{{xy = 3/2}}}<br>
And finally
{{{x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 27 - 3(3/2)(3) = 27-27/2 = 27/2}}}<br>
Another way to get an expression that includes x^3 and y^3 is to multiply the two given equations together.
{{{(x+y)(x^2+y^2) = x^3 + x^2y + xy^2 + y^3 = (x^3 + y^3) + xy(x+y)}}}
{{{x^3 + y^3 = (x+y)(x^2+y^2) - xy(x+y)}}}
{{{x^3 + y^3 = (3)(6) - (3/2)(3) = 18-9/2 = 27/2}}}