if the product of tree consecutive intergers is 120 then the sum of the integers is
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let the numbers be (x-1),(x), (x+1)
atq
(x-1).(x).(x+1)=120
=>(x^2-1)x=120
=>x^3-x=120
=>x^3-x-120=0
In the above eq. there can be three values for x among which 2 are imaginary and only1 is real. The real value is 5. The imaginary values are[(-5/2)-{(I*1)/2√71}],[(-5/2)+{(I*1)/2√71}]
Therefore considering only the real values, the numbers are 4,5,6. Thus the sum is 15
atq
(x-1).(x).(x+1)=120
=>(x^2-1)x=120
=>x^3-x=120
=>x^3-x-120=0
In the above eq. there can be three values for x among which 2 are imaginary and only1 is real. The real value is 5. The imaginary values are[(-5/2)-{(I*1)/2√71}],[(-5/2)+{(I*1)/2√71}]
Therefore considering only the real values, the numbers are 4,5,6. Thus the sum is 15
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