Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.

Given:

Let a be the side of each cube.

When the cubes are placed adjacently, they form a cuboid with the following dimensions:

Length of cuboid = a + a + a = 3a

Breadth of cuboid = a

Height of cuboid = a

The ratio = $\dfrac{\text{Tot. surface area of cuboid}}{\text{Sum of tot. surface areas of 3 cubes}}$

$= \dfrac{2(lb + bh + hl)}{6 \times side^2 + 6 \times side^2 + 6 \times side^2}\\[1em] = \dfrac{2(3a \times a + a \times a + a \times 3a)}{3 \times 6a^2}\\[1em] = \dfrac{2(3a^2 + a^2 + 3a^2)}{18a^2}\\[1em] = \dfrac{2 \times 7a^2}{18a^2}\\[1em] = \dfrac{14a^2}{18a^2}\\[1em] = \dfrac{14 \cancel{a^2}}{18 \cancel{a^2}}\\[1em] = \dfrac{14}{18}\\[1em] = \dfrac{7}{9}\\[1em]$

Hence, the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes is 7:9.