Question 1046508
let three consecutive integers be {{{ x}}},{{{(x+1)}}},and{{{(x+2)}}}

given: The product of three consecutive integers is {{{185136}}}.

so, we have 

{{{ x(x+1)(x+2)=185136}}} ...solve for {{{x}}}

{{{ (x^2+x)(x+2)=185136}}}

{{{ x^3+x^2+2x^2+2x=185136}}}

{{{ x^3+3x^2+2x-185136=0}}}...factor

{{{ x^3+59x^2-56x^2+3306x-56*59x-56*3306=0}}}

{{{ (x^3-56x^2)+(59x^2-56*59x)+(3306x-56*3306)=0}}}

{{{ x^2(x-56)+59x(x-56)+3306(x-56)=0}}}

{{{(x-56)(x^2+59x+3306) = 0}}}

real solution is: 
if{{{(x-56)= 0}}}->{{{highlight(x=56)}}}

{{{(x^2+59x+3306) = 0}}}...coefficient {{{c=3306}}} is positive and greater than {{{b=59}}} and we will have complex solutions; so, disregard it


first integer is {{{highlight(56)}}}
next one is {{{highlight(57)}}}
next one is {{{highlight(58)}}}

check their product: {{{56*57*58=185136}}} which is true

now find their sum: {{{56+57+58=171}}}


so, the sum of these integers is {{{highlight(171)}}}