Question 122902
{{{(a^-2-4b^-2)/(3b-6a)}}} Start with given expression



{{{(1/a^2-4/b^2)/(3b-6a)}}} Flip the bases that have negative exponents. Remember {{{a^-2=1/a^2}}}


{{{((b^2/b^2)(1/a^2)-(a^2/a^2)(4/b^2))/(3b-6a)}}} Multiply the first fraction {{{1/a^2}}} by {{{b^2/b^2}}}. Multiply the second fraction {{{1/b^2}}} by {{{a^2/a^2}}}. 



{{{((b^2)/(a^2b^2)-(4a^2)/(a^2b^2))/(3b-6a)}}} Multiply the fractions



{{{((b^2-4a^2)/(a^2b^2))/(3b-6a)}}} Combine the fractions in the numerator




{{{((b^2-4a^2)/(a^2b^2))*(1/(3b-6a))}}} Multiply the first fraction by the reciprocal of the second fraction




{{{(((b+2a)(b-2a))/(a^2b^2))*(1/(3(b-2a)))}}} Factor {{{b^2-4a^2}}} to get {{{(b+2a)(b-2a)}}}. Factor {{{3b-6a}}} to get {{{3(b-2a)}}}




{{{(((b+2a)cross((b-2a)))/(a^2b^2))*(1/(3*cross((b-2a))))}}} Cancel like terms




{{{(b+2a)/(3a^2b^2)}}} Simplify




Answer:



So {{{(a^-2-4b^-2)/(3b-6a)}}} simplifies to {{{(b+2a)/(3a^2b^2)}}}



In other words, {{{(a^-2-4b^-2)/(3b-6a)=(b+2a)/(3a^2b^2)}}}