$$z^3-2z^2+pz+10=0$$ $$ax^3+bx^2+cx+d=0$$ $$\Rightarrow\,\,\,\,\,\,\,\,\,a=1,\,\,\,\,\,\,\,\, b=-2,\,\,\,\,\,\,\,\, c=p,\,\,\,\,\,\,\,\, d=10$$
$$(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\alpha\gamma+\beta\gamma)$$ $$\Rightarrow\,\,\,\,\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma)$$ $$\alpha+\beta+\gamma=-\frac{b}{a}$$ $$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$$ $$\Rightarrow\,\,\,\,(-\frac{b}{a})^2-2(\frac{c}{a})$$ $$\Rightarrow\,\,\,\,(-\frac{(-2)}{(1)})^2-2(\frac{(p)}{(1)})$$ $$\Rightarrow\,\,\,\,4-2p$$
This is the only method I know.
Edit: Extra information if needed. $$\alpha^3+\beta^3+\gamma^3=-4$$
Given the roots of the equation, we have $$\begin{align}\alpha^3-2\alpha^2+p\alpha+10=0\\ \beta^3-2\beta^2+p\beta+10=0\\ \gamma^3-2\gamma^2+p\gamma+10=0\end{align}$$ Add all three equations and rearrange $$\alpha^2+\beta^2+\gamma^2=\frac{\alpha+\beta+\gamma}{2}p+\frac{\alpha^3+\beta^3+\gamma^3}{2}+15$$ Given the information already outlined in your question, I reckon you can take it from here.