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

Let edge of each equal cubes = x
Then, surface area of one cube = $6x^2$
and surface area of three cubes = $3\times 6x^2=18x^2$
By placing the cubes in a row,
The length of newly formed cubiod (l) = 3x
Breadth (b) = x
and height (h) = x
$\therefore$ Surface area of the cubiod so formed
= $2(l\times b+b\times h+h\times l)$
= $2(3x\times x+x\times x+x\times 3x)$
= $2(3x^2+x^2+3x^2)=2\times 7x^2=14x^2$
$\therefore$ Ratio = $14x^2:18x^2$ = 7:9