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x(x+1)(x+2) = x+(x+1)+(x+2)
x(x²+3x+2) = x+x+1+x+2
x³+3x²+2x = 3x+3
x³+3x²-x-3 = 0
x²(x+3)-1(x+3) = 0
(x+3)(x²-1) = 0
(x+3)(x-1)(x+1) = 0
x+3=0; x-1=0; x+1=0
x=-3 x=1; x=-1
The only one that's positive is x=1,
So the three consecutive integers are 1,2,3.
The sum of their squares = 1²+2²+3² = 1+4+9 = 14
Edwin