Answer:
[tex]\boxed{2^{\frac{802}{27}} \cdot 3^9}[/tex]
Step-by-step explanation:
I will try to give as many details as possible.
First of all, I just would like to say:
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[tex]$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$[/tex]
Note that
[tex]\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }[/tex]
The denominator can't be 0 because it would be undefined.
So, we can solve the expression inside both parentheses.
[tex]\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3[/tex]
Also,
[tex]\boxed{a^{0} = 1, a\neq 0 }[/tex]
[tex]\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27[/tex]
Note
[tex]\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}[/tex]
[tex]\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27[/tex]
[tex]\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)[/tex]
[tex]\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)[/tex]
[tex]\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)[/tex]
Note
[tex]\boxed{\dfrac{1}{\dfrac{1}{a} } = a}[/tex]
[tex]9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)[/tex]
[tex]\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)[/tex]
Once
[tex]9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6[/tex]
[tex]\boxed{(a \cdot b)^n=a^n \cdot b^n}[/tex]
And
[tex]$4^{\frac{4}{27}} = 2^{\frac{8}{27} $[/tex]
We have
[tex]\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)[/tex]
Also, once
[tex]\boxed{\dfrac{c^a}{c^b}=c^{a-b}}[/tex]
[tex]2^{30-\frac{8}{27}} \cdot 3^6\cdot 27[/tex]
As
[tex]30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}[/tex]
[tex]2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3[/tex]
[tex]2^{\frac{802}{27}} \cdot 3^9[/tex]
