If x
2
+9y
2
=40 and xy=4, where x>0,y>0, then what is the value of (x
3
+27y
3
)?

We can write (x+3y)
2
as:

(x+3y)
2
=x
2
+9y
2
+6xy
Substitute the given values:

(x+3y)
2
=40+6(4)=40+24=64
Since x>0,y>0, we have x+3y=8.

Now, we need to find x
3
+27y
3
:

x
3
+27y
3
=(x+3y)
3
−3⋅x⋅3y(x+3y)
x
3
+27y
3
=(x+3y)
3
−9xy(x+3y)
Substitute the values:

x
3
+27y
3
=(8)
3
−9(4)(8)
=512−288=224