How do you simplify #\frac { y ^ { 3} } { 14} \cdot \frac { 8} { 5y ^ { 3} }#?

1 Answer

#4/35#

Explanation:

since you have #y^3# in both numerator and denominator you can cancel them out.
#cancel(y^3)/14 * 8/(5*cancel y ^3)#
then you can factorize 2 from numerator and denominator and strike it out.
#1/(cancel2*7) * (2*2*cancel2)/(5)#
So your final answer is #4/35#