How do you simplify #(- x ^ { 2} y ^ { 3} z ^ { 3} ) ( - 2x ^ { 3} y ^ { 2} z ^ { 4} ) ^ { 3}#?

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Answer 1 · 2017-12-21T08:30:45

#8x^11y^9z^15#

We're going to use a lot of properties of exponents.

First we'll focus on the second set of parentheses: #(-2x^3y^2 z^4)^3# and use the property that #(a*b)^c=a^c*b^c#.

#(-2x^3y^2 z^4)^3 = (-2)^3(x^3)^3(y^2)^3(z^4)^3#

now we'll use the property that #(a^b)^c=a^(b*c)# and the fact that #(-2)^3=-8#:

#(-2)^3(x^3)^3(y^2)^3(z^4)^3=-8x^9y^6z^12#

so our original problem has reduced to:

#(-x^2y^3z^3)(-8x^9y^6z^12)#

now we'll use the property that #a^b*a^c=a^(b+c)#

#(-x^2y^3z^3)(-8x^9y^6z^12)=(-1)(-8)x^(2+9)y^(3+6)z^(3+12)#

#=8x^11y^9z^15#