Question 106576
{{{((-3a^2b)/(35a^5))*((14a^3b^2)/(-9b^4))}}}
Lets begin by canceling;
{{{(cross(-3)a^2b)/(cross(35)a^5)*(cross(14)a^3b^2)/(cross(-9)b^4)}}}
Now we have;
{{{((a^2b)/(5a^5))*((2a^3b^2)/(3b^4))}}}
Lets cancel more;
{{{(a^2cross(b))/(5cross(a^5))*(2cross(a^3)b^2)/(3cross(b^4))}}}
Now we have:
{{{((a^2)/(5a^2))*((2b^2)/(3b^3))=(2a^2b^2)/(15a^2b^3)}}}
Almost there, we can still cancel;
{{{(2cross(a^2)cross(b^2))/(15cross(a^2)cross(b^3))=(2)/(15b)}}}
Whew! That was long!
:)