How do you simplify #(\frac { 2x ^ { 4} y ^ { 0} z ^ { - 1} } { x ^ { - 1} y ^ { 2} \cdot x y ^ { - 4} z ^ { 4} } ) ^ { - 4}#?

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Answer 1 · 2017-02-02T14:44:00

#z^20/(16x^16y^8)#

#color(brown)("The trick with these is to take it very slowly and 1 step at a time" )#

Note that #y^0=1#

#color(blue)("Consider the numerator")#

#2x^4xx1xx1/z = (2x^4)/z#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the denominator" )#

#1/x xxy^2xx x xx1/y^4xxz^4#

But in fact this is:

#1/(1/x xxy^2xx x xx1/y^4xxz^4)" "=cancel(x) xx 1/(cancel(x)) xx 1/(cancel(y^2))xxy^(cancel(color(white)(.)4)2)xx1/z^4#

# =y^2/z^4 #

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all back together")#

Putting the numerator and denominator back together we have:

#(2x^4)/z xx y^2/z^4 = (2x^4y^2)/z^5#

But the whole thing is raised to the power of negative 4 giving

#((2x^4y^2)/(z^5))^(-4)#

#((z^5)/(2x^4y^2))^4#

#z^20/(16x^16y^8)#