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Advocard [28]
3 years ago
13

How do you do this. Please give a full answer and explain why.

Mathematics
1 answer:
OverLord2011 [107]3 years ago
3 0

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u>

\displaystyle \large{ {a}^{ - n}  =  \frac{1}{ {a}^{n} } }

Compare the terms.

\displaystyle \large{ {a}^{ - n}  =   {( - 2)}^{ - 3} }

Therefore, a = -2 and n = 3. From the law of exponent above, we receive:

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{ {( - 2)}^{ 3} } }

<u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>D</u><u>e</u><u>f</u><u>.</u> (For cubic)

\displaystyle \large{ {a}^{3}  = a \times a \times a }

Factor (-2)^3 out.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{( - 2) \times ( - 2) \times ( - 2)}}

(-2) • (-2) = 4 | Negative × Negative = Positive.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{4 \times ( - 2)}}

4 • (-2) = -8 | Negative Multiply Positive = Negative.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{  - 8}}

If either denominator or numerator is in negative, it is the best to write in the middle or between numerator and denominators.

Hence,

\displaystyle \large \boxed{ {( - 2)}^{ - 3}  =  -  \frac{1}{  8}}

The answer is - 1 / 8

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