What is (-3p^{-3}q*3p^3q^{-4})^2 ?

The solution to (-3p^{-3}q*3p^3q^{-4})^2 is (81)/(q^6)

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simplify (-3p^{-3}q*3p^3q^{-4})^2

simplify

(- 3 p^{- 3} q \cdot 3 p^{3} q^{- 4})^{2}

\frac{81}{q ^{6}}

Solution steps

(- 3 p^{- 3} q \cdot 3 p^{3} q^{- 4})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3 p^{- 3} q \cdot 3 p^{3} q^{- 4})^{2} = (3 p^{- 3} q \cdot 3 p^{3} q^{- 4})^{2}

= (3 p^{- 3} q \cdot 3 p^{3} q^{- 4})^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(3 p^{- 3} q \cdot 3 p^{3} q^{- 4})^{2} = 3^{2} (p^{- 3})^{2} q^{2} \cdot 3^{2} (p^{3})^{2} (q^{- 4})^{2}

= 3^{2} (p^{- 3})^{2} q^{2} \cdot 3^{2} (p^{3})^{2} (q^{- 4})^{2}

3^{2} = 9

= 9 (p^{- 3})^{2} q^{2} \cdot 3^{2} (p^{3})^{2} (q^{- 4})^{2}

Simplify

(p^{- 3})^{2} : \frac{1}{p ^{6}}

= 9 \cdot \frac{1}{p ^{6}} q^{2} \cdot 3^{2} (p^{3})^{2} (q^{- 4})^{2}

3^{2} = 9

= 9 \cdot \frac{1}{p ^{6}} q^{2} \cdot 9 (p^{3})^{2} (q^{- 4})^{2}

Simplify

(p^{3})^{2} : p^{6}

= 9 \cdot \frac{1}{p ^{6}} q^{2} \cdot 9 p^{6} (q^{- 4})^{2}

Simplify

(q^{- 4})^{2} : \frac{1}{q ^{8}}

= 9 \cdot \frac{1}{p ^{6}} q^{2} \cdot 9 p^{6} \frac{1}{q ^{8}}

Multiply fractions:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

= \frac{1 \cdot 1 \cdot 9 q ^{2} \cdot 9 p ^{6}}{p ^{6} q ^{8}}

Cancel the common factor:

p^{6}

= \frac{1 \cdot 1 \cdot 9 q ^{2} \cdot 9}{q ^{8}}

Multiply the numbers:

1 \cdot 1 \cdot 9 \cdot 9 = 81

= \frac{81 q ^{2}}{q ^{8}}

Apply exponent rule:

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

\frac{q ^{2}}{q ^{8}} = \frac{1}{q ^{8 - 2}}

= \frac{81}{q ^{8 - 2}}

Subtract the numbers:

8 - 2 = 6

= \frac{81}{q ^{6}}

x^{2} + 3 x + 7 > 0

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

\frac{5 x ^{2} y ^{8}}{5 x ^{4} y ^{8}}

1^{2} + x^{2} = 2^{2}

x^{2} + 18 x = - 80

What is (-3p^{-3}q*3p^3q^{-4})^2 ?
The solution to (-3p^{-3}q*3p^3q^{-4})^2 is (81)/(q^6)