Question 224739
# 1



{{{((3)/(a^2+a))((2a+2)/(6))}}} Start with the given expression.



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



{{{((3)/(a(a+1)))((2(a+1))/(6))}}} Factor {{{2a+2}}} to get {{{2(a+1)}}}.



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



{{{(3*2(a+1))/(a*(a+1)(2*3))}}} Combine the fractions. 



{{{(highlight(3)highlight(2)highlight((a+1)))/((a)highlight((a+1))(highlight(2)*highlight(3)))}}} Highlight the common terms. 



{{{(cross(3)cross(2)cross((a+1)))/((a)cross((a+1))(cross(2)*cross(3)))}}} Cancel out the common terms. 



{{{1/a}}} Simplify. 



So {{{((3)/(a^2+a))((2a+2)/(6))}}} simplifies to {{{1/a}}}.



In other words, {{{((3)/(a^2+a))((2a+2)/(6))=1/a}}} where {{{a<>0}}} or {{{a<>-1}}}



<hr>



# 2





{{{((w^2-1)/((w-1)^2))((w-1)/(w^2+2w+1))}}} Start with the given expression.



{{{(((w-1)*(w+1))/((w-1)^2))((w-1)/(w^2+2w+1))}}} Factor {{{w^2-1}}} to get {{{(w-1)*(w+1)}}}.



{{{(((w-1)*(w+1))/((w-1)(w-1)))((w-1)/(w^2+2w+1))}}} Break up {{{(w-1)^2}}} to get {{{(w-1)(w-1)}}}.



{{{(((w-1)*(w+1))/((w-1)(w-1)))((w-1)/((w+1)(w+1)))}}} Factor {{{w^2+2w+1}}} to get {{{(w+1)(w+1)}}}.



{{{((w-1)*(w+1)(w-1))/((w-1)(w-1)(w+1)(w+1))}}} Combine the fractions. 



{{{(highlight((w-1))highlight((w+1))highlight((w-1)))/(highlight((w-1))highlight((w-1))highlight((w+1))(w+1))}}} Highlight the common terms. 



{{{(cross((w-1))cross((w+1))cross((w-1)))/(cross((w-1))cross((w-1))cross((w+1))(w+1))}}} Cancel out the common terms. 



{{{1/(w+1)}}} Simplify. 



So {{{((w^2-1)/((w-1)^2))((w-1)/(w^2+2w+1))}}} simplifies to {{{1/(w+1)}}}.



In other words, {{{((w^2-1)/((w-1)^2))((w-1)/(w^2+2w+1))=1/(w+1)}}}