surface area of cube

Problem 01
Show that (a) the total surface of the cube is twice the square of its diagonal, (b) the volume of the cube is $\frac{1}{9}\sqrt{3}$ times the cube of its diagonal.
 

Solution 01
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!