Question 678239
{{{((81b^2)/(4b^2-24b+36))*((4b-12)/(9b))}}}

Simplify on the diagonal {{{81b^2}}} with {{{9b}}}:

{{{((9b)/(4b^2-24b+36))*((4b-12)/(1))}}}

Factor out a 4 both denominator of first fraction and numerator of second fraction:

{{{((9b)/(4(b^2-6b+9)))*((4(b-3))/(1))}}}

Simplify the 4's that you just factor out:

{{{(9b/(b^2-6b+9))*((b-3)/1)}}}

Factor the quadratic {{{b^2-6b+9 = (b-3)^2}}}:

{{{(9b/(b-3)^2)*((b-3)/1)}}}

Simplify {{{(b-3)^2}}} with {{{b-3}}}:

{{{(9b/(b-3))*(1/1)}}}

Result:

{{{9b/(b-3)}}}