How do you multiply $(2a ^ { - 2} ) ^ { 3} ( a ^ { 4} a ^ { - 5} ) ^ { 6}$ ?

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Answer 1 · 2018-07-10T07:26:51

$8/a^(12)$

We have $(2a^(-2))^3(a^4*a^(-5))^6$
Note that $a^4*a^(-5)=a^(-1)$
and $(ab)^n=a^n*b^n$ so we get

$2^3*a^(-6)*a^(-6)=8*a^(-12)=8/a^12$

Answer 2 · 2018-07-10T07:30:40

$8/a^12$

Given tha

$(2a^{-2})^3(a^4a^{-5})^6$

$=2^3a^{-2\cdot 3}(a^{-5+4})^6$

$=8a^{-6}(a^{-1})^6$

$=8a^{-6}\cdot a^{-6}$

$=8a^{-6-6}$

$=8a^{-12}$

$=8/a^12$

How do you multiply (2a ^ { - 2} ) ^ { 3} ( a ^ { 4} a ^ { - 5} ) ^ { 6}(2a ^ { - 2} ) ^ { 3} ( a ^ { 4} a ^ { - 5} ) ^ { 6}?