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

The solution to (((-2a^3b^2c^0))/(3a^2b^3c^7))^{-2} is (9b^2c^{14})/(4a^2)

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

simplify

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

\frac{9 b ^{2} c ^{14}}{4 a ^{2}}

Solution steps

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

Apply exponent rule:

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

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

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

Simplify

(\frac{- 2 a ^{3} b ^{2} c ^{0}}{3 a ^{2} b ^{3} c ^{7}})^{2} : \frac{4 a ^{2}}{9 b ^{2} c ^{14}}

= \frac{1}{\frac{4 a ^{2}}{9 b ^{2} c ^{14}}}

Apply the fraction rule:

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

= \frac{9 b ^{2} c ^{14}}{4 a ^{2}}

\frac{3 x ^{4} - 2 x ^{2} - 5}{3 x ^{2} - 5}

(2 \cdot 6)^{\frac{3}{2}}

1 6^{- 4^{- 2^{- 1}}}

(x + 1)^{- 2}

\frac{2 + 6}{6 + 8}

What is (((-2a^3b^2c^0))/(3a^2b^3c^7))^{-2} ?
The solution to (((-2a^3b^2c^0))/(3a^2b^3c^7))^{-2} is (9b^2c^{14})/(4a^2)