prove that the square of any whole number is 1 greater than the product of the preceding and following whole numbers.

Let x be the whole number. x^2 is the square of the number. x-1 is the preceding number. And x+1 is the following number.

(x+1)(x-1)=

x^2 + x - x - 1

x^2 -1

Your answer is x^2-1 which is one less that x^2. This also means that the square of any whole number (x^2) is one greater than the product of the preceding (x-1) and the following (x+1).

x/3-10=8

x/3=18

x=48