x, x+1, x+2 are the consecutive integers product of which is equal to 336

x * x +1 * x + 2 = x (x^2 + 3x + 2) = x^3 + 3 x^2 + 2x = 336

x^3 + 3 x^2 + 2x - 336 = 0

From here you have to test potential roots of f(x) = x^3 + 3 x^2 + 2x - 336. You can use prime factorization to start. Normal suspects of x = 2 or x = 3 did not work, so I expanded the options.

If f(root) = 0 then root is a solution. In this situation f(6) = 0 which implies (x - 6) is a factor of x^3 + 3 x^2 + 2x - 336. Using polynominal division (or synthetic division) we find the factors of x^3 + 3 x^2 + 2x - 336 are (x-6) and (x^2 +9x +56).
since the only integer solutions to (x-6) (x^2 +9x +56) = 0 is x = 6, we surmise the three consecutive integers are 6, 7 and 8 (which double checking is equal to 336)

Hope this helps.

Regards,

Alvin

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