What is (ab^2c^{-3})^{-2\circ}(ab^2c) ?

The solution to (ab^2c^{-3})^{-2\circ}(ab^2c) is (c^{6\circ+1})/(ab^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (ab^2c^{-3})^{-2\circ}(ab^2c)

simplify

(a b^{2} c^{- 3})^{- 2 \circ} (a b^{2} c)

\frac{c ^{6 \circ+ 1}}{a b ^{2}}

Solution steps

(a b^{2} c^{- 3})^{- 2 \circ} (a b^{2} c)

Apply exponent rule:

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

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

\frac{1}{( a b ^{2} c ^{- 3} ) ^{2 \circ}} = \frac{c ^{6 \circ}}{a ^{2 \circ} b ^{4 \circ}}

= a b^{2} c \frac{c ^{6 \circ}}{a ^{2 \circ} b ^{4 \circ}}

Multiply fractions:

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

= \frac{c ^{6 \circ} a b ^{2} c}{a ^{2 \circ} b ^{4 \circ}}

c^{6 \circ} a b^{2} c = a b^{2} c^{6 \circ + 1}

= \frac{a b ^{2} c ^{6 \circ + 1}}{a ^{2 \circ} b ^{4 \circ}}

Cancel the common factor:

a

= \frac{b ^{2} c ^{6 \circ + 1}}{a b ^{4 \circ}}

Apply exponent rule:

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

\frac{b ^{2}}{b ^{4 \circ}} = \frac{1}{b ^{4 \circ - 2}}

= \frac{c ^{6 \circ + 1}}{a b ^{4 \circ - 2}}

Subtract the numbers:

4 \circ - 2 = 2

= \frac{c ^{6 \circ + 1}}{a b ^{2}}

4.620981204 \times 1 0^{- 5}

- 65^{2}

\frac{x ^{n}}{2 ^{n}}

What is (ab^2c^{-3})^{-2\circ}(ab^2c) ?
The solution to (ab^2c^{-3})^{-2\circ}(ab^2c) is (c^{6\circ+1})/(ab^2)