Certainly! Let's solve for [tex]\(x^3 + y^3\)[/tex] given the equations [tex]\(x + y = 4\)[/tex] and [tex]\(xy = 5\)[/tex].
First, recall the identity for the sum of cubes:
[tex]\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \][/tex]
To use this identity, we need to determine [tex]\(x^2 + y^2\)[/tex]. We start by squaring the sum [tex]\(x + y\)[/tex]:
[tex]\[ (x + y)^2 = x^2 + 2xy + y^2 \][/tex]
Since we know [tex]\(x + y = 4\)[/tex] and [tex]\(xy = 5\)[/tex], we substitute these values into the equation:
[tex]\[ 4^2 = x^2 + 2 \cdot 5 + y^2 \][/tex]
[tex]\[ 16 = x^2 + 10 + y^2 \][/tex]
Now, isolate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ 16 = x^2 + y^2 + 10 \][/tex]
[tex]\[ x^2 + y^2 = 16 - 10 \][/tex]
[tex]\[ x^2 + y^2 = 6 \][/tex]
Next, plug the values into the sum of cubes identity, [tex]\( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \)[/tex]:
[tex]\[ x^2 - xy + y^2 = x^2 + y^2 - xy \][/tex]
Using the values we have:
[tex]\[ x^2 + y^2 = 6 \quad \text{and} \quad xy = 5 \][/tex]
[tex]\[ x^2 - xy + y^2 = 6 - 5 = 1 \][/tex]
Finally, substitute these results into the expression for [tex]\( x^3 + y^3 \)[/tex]:
[tex]\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \][/tex]
[tex]\[ x + y = 4 \quad \text{and} \quad x^2 - xy + y^2 = 1 \][/tex]
[tex]\[ x^3 + y^3 = 4 \cdot 1 = 4 \][/tex]
Thus, the value of [tex]\( x^3 + y^3 \)[/tex] is [tex]\( \boxed{4} \)[/tex].
