How do you simplify ${(\frac{w^{4}}{3})}^{- 3}$ $(\frac { w ^ { 4} } { 3} ) ^ { - 3}$ ?

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Answer 1 · 2018-04-21T17:55:59

${(\frac{w^{4}}{3})}^{- 3} = \frac{27}{w^{12}}$ $(w^4/3)^-3=27/w^12$

In order to simplify ${(\frac{w^{4}}{3})}^{- 3}$ $(w^4/3)^-3$

Begin by distributing the exponent outside the parenthesis to both the numerator and the denominator

$(\frac{w^{(4) (- 3)}}{3^{- 3}}$ $((w^((4)(-3)))/3^-3$

This gives us $(\frac{w^{- 12}}{3^{- 3}})$ $(w^-12/3^-3)$

In order to eliminate the negative exponents invert numerator and the denominator

$\frac{3^{3}}{w^{12}}$ $3^3/w^12$

Now simplify the exponent value in the numerator $3^{3} = 27$ $3^3 = 27$

$\frac{27}{w^{12}}$ $27/w^12$

Answer 2 · 2018-04-21T18:30:26

$= \frac{27}{w^{12}}$ $=27/w^12$

By applying the laws of indices:

${(\frac{a}{b})}^{- m} = {(\frac{b}{a})}^{m}$ $(a/b)^-m = (b/a)^(m)$

${(\frac{w^{4}}{3})}^{- 3} = {(\frac{3}{w^{4}})}^{3}$ $(w^4/3)^-3 = (3/w^4)^3$

Now raising each factor in the bracket to the power of $3$ $3$ gives:

$\frac{3^{3}}{w^{4 \times 3}}$ $3^3/w^(4xx3)$

$= \frac{27}{w^{12}}$ $=27/w^12$

How do you simplify (w43)−3(\frac { w ^ { 4} } { 3} ) ^ { - 3}?