How do you simplify ${(w^{- 5})}^{3}$ $(w ^ { - 5} ) ^ { 3}$ without using negative exponents?

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Answer 1 · 2017-04-15T15:28:36

$\frac{1}{w^{15}}$ $1/w^15$

We will use two of the laws of powers

$(i) {(x^{m})}^{n} = x^{m n}$ $(i)" " (x^m)^n=x^(mn)$

$(i i) x^{- m} = \frac{1}{x^{m}}$ $(ii)" "x^(-m)=1/(x^m)$

${(w^{- 5})}^{3} = w^{- 5 \times 3} = w^{- 15}$ $(w^(-5))^3=w^(-5xx3)=w^(-15)$

$= \frac{1}{w^{15}}$ $=1/w^15$

How do you simplify (w−5)3(w ^ { - 5} ) ^ { 3} without using negative exponents?