The product of 3 consecutive positive integers is equal to 56 more than the cube of the first integer. What is largest of these integers?

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If the first integer is x, then the second integer is (x + 1) and the third integer is (x + 2). The product of these integers is x(x + 1)(x + 2); x(x + 1) = x 2 + x; (x 2 + x)(x + 2) = x 3 + 2x 2 + x 2 + 2x = x 3 + 3x 2 + 2x. This polynomial is equal to the cube of the first integer, x 3 , plus 56. Therefore, x 3 + 3x 2 + 2x = x 3 + 56. Place all terms on one side of the equation, combining like terms; x 3 + 3x 2 + 2x = x 3 + 56, 3x 2 + 2x – 56 = 0. Factor the polynomial: 3x 2 + 2x – 56 = (3x + 14)(x – 4). Set each term equal to 0 to find the values of x that make the equation true; 3x + 14 = 0, 3x = –14, x = – 14 ⁄ 3 , but you are looking for a positive integer; x – 4 = 0, x = 4. If 4 is the first of the consecutive integers, then 5 is the second integer and 6 is the third, and largest, integer.

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