What is ((((4x^3y^{-2})^3)^0)^5)^{-4} ?

The solution to ((((4x^3y^{-2})^3)^0)^5)^{-4} is 1

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simplify ((((4x^3y^{-2})^3)^0)^5)^{-4}

simplify

((((4 x^{3} y^{- 2})^{3})^{0})^{5})^{- 4}

1

Solution steps

((((4 x^{3} y^{- 2})^{3})^{0})^{5})^{- 4}

((4 x^{3} y^{- 2})^{3})^{0} : (4 x^{3} y^{- 2})^{0}

= (((4 x^{3} y^{- 2})^{0})^{5})^{- 4}

Apply rule

a^{0} = 1, a \neq = 0

(4 x^{3} y^{- 2})^{0} = 1

= (1^{5})^{- 4}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{( 1 ^{5} ) ^{4}}

(1^{5})^{4} : 1^{20}

= \frac{1}{1 ^{20}}

Apply rule

1^{a} = 1

1^{20} = 1

= \frac{1}{1}

Apply the fraction rule:

\frac{a}{1} = a

= 1

(\frac{1}{3})^{3}

(5 x^{2}) (2 x^{4})

(36)^{2}

(x + 7)^{2}

5.4 \cdot 1 0^{- 3}

What is ((((4x^3y^{-2})^3)^0)^5)^{-4} ?
The solution to ((((4x^3y^{-2})^3)^0)^5)^{-4} is 1