Final answer:
To find x^3-y^3 while given x-y=4 and x^2+y^2=5, we used the system of equations and algebraic identities. After deriving xy=11/2 by manipulating the given equations, we used the identity for the difference of cubes to find the final solution, which is x^3-y^3=42.
Step-by-step explanation:
To find the value of x^3-y^3 given that x-y=4 and x^2+y^2=5, we will use the system of equations and algebraic identities. First, by adding the two equations, we get (x+y)2, as:
(x+y)2 = x2+2xy+y2 = (x2+y2) + 2xy
5 + 2xy = (x+y)2
We have x-y=4, hence (x+y) can be expressed in terms of x:
x+y = x-(x-4) = 4
Thus, (4)2 = 5 + 2xy
16 = 5 + 2xy
2xy = 11
xy = 11/2
Now we use the identity x^3-y^3 = (x-y)(x2+xy+y2):
x^3-y^3 = 4(x2+xy+y2)
Finally, substituting the known values:
x^3-y^3 = 4(5 + 11/2)
x^3-y^3 = 4(10 + 11)/2
x^3-y^3 = 42