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Let x , x+1 and x+2 be the consecutive numbers

x(x+1)(x+2) = 1,320

x(x²+3x+2) = 1,320
x³+3x²+2x = 1,320
x³+3x²+2x-1,320 = 0
This is really difficult to factor because 1,320 has many factors, and the only way to find the factors is by trial and error. So, it would be easier to just guess three consecutive numbers that will satisfy the problem.

The clue here is that 1,320 ends in 0, so it must have a factor that ends in 0. 10·10·10 = 1,000 , and this is really close to 1,320, thus, we get the next 2 numbers after 10.

Therefore, the three numbers are 10, 11, 12.

0 0 0
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