$\displaystyle \alpha^3 - p\alpha^2 + \beta^3 - p\beta^2 =( \alpha^3 + \beta^3) -p (\alpha^2 + \beta^2) $

$\displaystyle =(\alpha + \beta)(\alpha^2 + \beta^2-\alpha \beta)-p\{(\alpha + \beta)^2 -2 \alpha \beta \}$

$\displaystyle \color {blue} \{ \because a^3+b^3 =(a+b)(a^2-ab+b^2) \quad and \quad a^2+b^2 =(a+b)^2-2ab \}$

$\displaystyle =(\alpha + \beta) \{(\alpha + \beta)^2-2 \alpha \beta -\alpha \beta \} -p\{(\alpha + \beta)^2 -2 \alpha \beta \}$

$\displaystyle = (\alpha + \beta) \{(\alpha + \beta)^2-3 \alpha \beta \} -p\{(\alpha + \beta)^2 -2 \alpha \beta \}$

substituting values of $\displaystyle (\alpha + \beta) \ and \ \alpha \beta $ from (2) and (3)

$\displaystyle = p \{ (p)^2-3q \} -p \{(p)^2-2q \}= p^3-3pq-p^3+2pq =-pq$