Question 1022393
{{{8x^3-125=0}}}



{{{(2x)^3-125=0}}}



{{{(2x)^3-5^3=0}}}



{{{(2x-5)((2x)^2+(2x)*5+5^2)=0}}} Use the difference of cubes factoring formula



{{{(2x-5)(4x^2+10x+25)=0}}}



{{{2x-5=0}}} or {{{4x^2+10x+25=0}}}



Solving {{{2x-5=0}}} yields {{{x = 5/2=2.5}}}



Solving {{{4x^2+10x+25=0}}} produces NO real solutions (the two solutions are imaginary). Use the quadratic formula to determine this.



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So the only solution is {{{x = 5/2}}} or {{{x = 2.5}}}. The two are equivalent but in different forms.