Question 880568
Since {{{a^3b^5c^7<0}}} , we know that {{{a^3b^5c^7<>0}}} , so {{{system(a<>0,b<>0,"and",c<>0)}}} .
That is good to know, in case we want to divide by those numbers.
For all {{{a<>0}}} and {{{c<>0}}} , {{{a^2>0}}} and {{{c^6=(c^3)^2>0}}} , because they are squares.
If {{{b>0}}} , then {{{b^5>0}}} .
So {{{a^2}}} , {{{b^5}}} and {{{c^6}}} are positive, and so is their product, {{{a^2b^5c^6}}} .
Then, starting with
{{{a^3b^5c^7<0}}} ,
we can divide both sides of the inequality by that positive product:
{{{a^3b^5c^7/a^2b^5c^6}}}{{{"<"}}}{{{0/a^2b^5c^6}}}
That simplifies to {{{highlight(ac<0)}}}
 
NOTE:
Of course, {{{a^2c^2>0}}} also must be true if {{{system(a<>0,"and",c<>0)}}} , which is a consequence of {{{a^3b^5c^7<0}}} .
So for  {{{a^3b^5c^7<0}}} , {{{a^2c^2>0}}} is true if {{{b>0}}} , and it is also true if {{{b<0}}} ,
but III is not the answer they want,
and the answer that will help your grade is {{{highlight(II)}}} .