$\displaystyle 6.c,d$
$\displaystyle \begin{array}{l}\frac{{dV}}{{dx}} = 12{x^2} - 52x + 40\\12{x^2} - 52x + 40 = 0 \Rightarrow 3{x^2} - 13x + 10 = 0\\{x_1} = \frac{{10}}{3};{x_2} = 1\\{V_{\max }} = 4 \cdot {\left( {\frac{{10}}{3}} \right)^3} - 26 \cdot {\left({\frac{{10}}{3}} \right)^2} + 40 \cdot \frac{{10}}{3}\end{array}$