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

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

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

simplify

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

4

Solution steps

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

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

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

Apply rule

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

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

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

Apply rule

1^{a} = 1

1^{5} = 1

= 4 \cdot 1^{-}

Apply rule

1^{a} = 1

1^{-} = 1

= 1 \cdot 4

Multiply the numbers:

1 \cdot 4 = 4

= 4

e^{- 2 x} \cdot e^{- 2 x}

e^{- 5 t}

(x^{2} - x - 2) (x^{2} - x - 2)

3 x^{2} (4 x^{2} y^{4})

x^{2} y^{3} (16 x^{3} y^{4})

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