and \( x y=21 \) find \( x^{3}+y^{3} \) \( x^{3}+y^{3}=(x+y) \)

To find \( x^3 + y^3 \), we can use the identity \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \). Given that \( xy = 21 \), we need to find \( x + y \) first. Let's denote \( s = x + y \) and \( p = xy \). The quadratic equation formed is \( t^2 - st + p = 0 \). The roots will yield \( x \) and \( y \). Once we have \( s \), we can compute \( x^2 + y^2 \) using \( x^2 + y^2 = (x + y)^2 - 2xy = s^2 - 2p \). Then we can easily substitute these values into the identity to solve for \( x^3 + y^3 \)! Now let's say \( s = 7 \) (for example), then \( x^2 + y^2 = 7^2 - 2(21) = 49 - 42 = 7 \). Thus, \( x^3 + y^3 = 7(7 - 21 + 7) = 7(7 - 21 + 7) = 7(-7) = -49 \). If you have actual values for \( x \) and \( y \) that fit \( xy = 21 \), plug those into the formulas to calculate \( x^3 + y^3 \)!