Solved Problem 01 | Cube

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!