Question 120541

{{{((2c^2-8)/(4c^2-8c))/((c^3+2c^2)/(8c^3))}}} Start with the given expression



{{{((2c^2-8)/(4c^2-8c))((8c^3)/(c^3+2c^2))}}} Multiply the first fraction by the reciprocal of the second fraction


{{{((2(c+2)(c-2))/(4c^2-8c))((8c^3)/(c^3+2c^2))}}}   Factor {{{2c^2-8}}} to get {{{2(c+2)(c-2)}}} 


{{{((2(c+2)(c-2))/(4c(c-2)))((8c^3)/(c^3+2c^2))}}}   Factor {{{4c^2-8c}}} to get {{{4c(c-2)}}} 


{{{((2(c+2)(c-2))/(4c(c-2)))((8c^2*c)/(c^3+2c^2))}}}   Factor {{{8c^3}}} to get {{{8c^2*c}}} 


{{{((2(c+2)(c-2))/(4c(c-2)))((8c^2*c)/(c^2(c+2)))}}}   Factor {{{c^3+2c^2}}} to get {{{c^2(c+2)}}} 



{{{2(c+2)(c-2)(8c^2*c)/(4c(c-2)c^2(c+2))}}} Combine the fractions



{{{2cross((c+2))cross((c-2))(8cross(c^2)*c)/(4c*cross((c-2))cross(c^2)cross((c+2)))}}} Cancel like terms




{{{(2*8c)/(4c)}}} Simplify




{{{16c/(4c)}}} Multiply




{{{4}}} Divide




So {{{((2c^2-8)/(4c^2-8c))((8c^3)/(c^3+2c^2))}}} simplifies to {{{4}}}




In other words, {{{((2c^2-8)/(4c^2-8c))((8c^3)/(c^3+2c^2))=4}}}