How do you simplify #sqrt(x^3y^3 )/sqrt(xy)#?

1 Answer
Zor Shekhtman
Oct 1, 2015

#sqrt(x^3*y^3)/sqrt(x*y) = x*y#
for all #x != 0# and #y != 0#
and it's undefined if either #x=0# or #y=0#

Explanation:

It's essential, before transforming any algebraic expression, to determine its domain, because during transformations we might derive with seemingly equivalent expression that has a different domain, and we will not have the right to say that original and final expressions are equivalent.

In this case we should exclude values #x=0# and #y=0# as those, when the expression is undefined since its denominator would by #0#.

For all other cases, when #x != 0# and #y != 0# we transform the expression as follows:

#sqrt(x^3*y^3)/sqrt(x*y) = sqrt(x^2*y^2*x*y)/sqrt(x*y) = #
#= sqrt(x^2*y^2)*sqrt(x*y)/sqrt(x*y) = #
# = sqrt((x*y)^2)*sqrt(x*y)/sqrt(x*y) = #
#= x*y*sqrt(x*y)/sqrt(x*y) = x*y*1 = x*y#

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