Question 153091
{{{((3k)/(3k^2-7k))((3k^3-7k^2)/(3k^3-6k))}}} Start with the given expression.



{{{((3k)/(k(3k-7)))((3k^3-7k^2)/(3k^3-6k))}}} Factor {{{3k^2-7k}}} to get {{{k*(3k-7)}}}.



{{{((3k)/(k(3k-7)))((k^2(3k-7))/(3k^3-6k))}}} Factor {{{3k^3-7k^2}}} to get {{{k^2(3k-7)}}}.



{{{((3k)/(k(3k-7)))((k^2(3k-7))/(3k(k^2-2)))}}} Factor {{{3k^3-6k}}} to get {{{3k(k^2-2)}}}.



{{{(3k*k^2(3k-7))/(3k^2(3k-7)(k^2-2))}}} Combine the fractions. 



{{{(highlight(3)k*highlight(k^2)*highlight((3k-7)))/(highlight(3)highlight(k^2)*highlight((3k-7))(k^2-2))}}} Highlight the common terms. 



{{{(cross(3)k*cross(k^2)*cross((3k-7)))/(cross(3)cross(k^2)*cross((3k-7))(k^2-2))}}} Cancel out the common terms. 



{{{(k)/(k^2-2)}}} Simplify. 



So {{{((3k)/(3k^2-7k))((3k^3-7k^2)/(3k^3-6k))}}} simplifies to {{{(k)/(k^2-2)}}}.



In other words, {{{((3k)/(3k^2-7k))((3k^3-7k^2)/(3k^3-6k))=(k)/(k^2-2)}}} where {{{k<>-2}}}, {{{k<>0}}}, {{{k<>2}}}, or {{{k<>7/3}}}