x + y = 3 => (x + y)² = 9 <=> x² + y² + 2xy = 9 <=> 5 + 2xy = 9 (because x² + y² = 5)

<=> 2xy = 4 <=> xy = 2

We have: (x + y)(x² + y²) = 3.5

<=> x³ + x²y + xy² + y³ = 15

<=> x³ + y³ + xy(x + y) = 15

<=> x³ + y³ + 2.3 = 15 (because xy = 2; x + y = 3)

<=> x³ + y³ + 6 = 15

<=> x³ + y³ = 9

So x³ + y³ = 9