If $V$ is the volume of a cuboid of dimensions $a, b, c$ and $S$ is its total surface area,then prove that $\frac{1}{V} = \frac{2}{S} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$.

(N/A) The volume $V$ of a cuboid with dimensions $a, b, c$ is given by $V = abc$.
The total surface area $S$ of the cuboid is given by $S = 2(ab + bc + ca)$.
Now,consider the right-hand side $(RHS)$ of the expression:
$RHS = \frac{2}{S} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$
Substitute the expression for $S$:
$RHS = \frac{2}{2(ab + bc + ca)} \left( \frac{bc + ac + ab}{abc} \right)$
Simplify the expression:
$RHS = \frac{1}{ab + bc + ca} \cdot \frac{ab + bc + ca}{abc}$
Cancel the common term $(ab + bc + ca)$:
$RHS = \frac{1}{abc}$
Since $V = abc$,we have:
$RHS = \frac{1}{V}$
Thus,$\frac{1}{V} = \frac{2}{S} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$ is proved.