What is ((-7a^2b^3c^0)/(3a^3b^4c^3))^{-4} ?

The solution to ((-7a^2b^3c^0)/(3a^3b^4c^3))^{-4} is (81a^4b^4c^{12})/(2401)

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simplify ((-7a^2b^3c^0)/(3a^3b^4c^3))^{-4}

simplify

(\frac{- 7 a ^{2} b ^{3} c ^{0}}{3 a ^{3} b ^{4} c ^{3}})^{- 4}

\frac{81 a ^{4} b ^{4} c ^{12}}{2401}

Solution steps

(\frac{- 7 a ^{2} b ^{3} c ^{0}}{3 a ^{3} b ^{4} c ^{3}})^{- 4}

Apply exponent rule:

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

(\frac{- 7 a ^{2} b ^{3} c ^{0}}{3 a ^{3} b ^{4} c ^{3}})^{- 4} = \frac{1}{( \frac{- 7 a ^{2} b ^{3} c ^{0}}{3 a ^{3} b ^{4} c ^{3}} ) ^{4}}

= \frac{1}{( \frac{- 7 a ^{2} b ^{3} c ^{0}}{3 a ^{3} b ^{4} c ^{3}} ) ^{4}}

Simplify

(\frac{- 7 a ^{2} b ^{3} c ^{0}}{3 a ^{3} b ^{4} c ^{3}})^{4} : \frac{2401}{81 a ^{4} b ^{4} c ^{12}}

= \frac{1}{\frac{2401}{81 a ^{4} b ^{4} c ^{12}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{81 a ^{4} b ^{4} c ^{12}}{2401}

(x + 3) (x - 2) \leq 0

2 x^{2} + 18 = 0

∣ x - 1 ∣ > 3

3 t - \frac{1}{4} = 0.75

x - \frac{3}{3} + 9 x = 7 x - 9 + (- 7 + 7)

What is ((-7a^2b^3c^0)/(3a^3b^4c^3))^{-4} ?
The solution to ((-7a^2b^3c^0)/(3a^3b^4c^3))^{-4} is (81a^4b^4c^{12})/(2401)