Space diagonal $s = a\sqrt{3}$, thus, $a = \dfrac{s}{\sqrt{3}}$
(a) Show that A = 2s2
$A = 6a^2$
$A = 6\left( \dfrac{s}{\sqrt{3}} \right)^2$
$A = 6\left( \dfrac{s^2}{3} \right)$
$A = 2s^2$ okay!
(b) Show that V = s3sqrt(3)/9
$V = a^3$
$V = \left( \dfrac{s}{\sqrt{3}} \right)^3$
$V = \frac{1}{9}\sqrt{3}\,s^3$ okay!