How do you simplify ${(\frac{2 x^{- 6}}{7 y^{- 3}})}^{- 3}$ $((2x^-6)/(7y^-3))^-3$ ?

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How do you simplify ${(\frac{2 x^{- 6}}{7 y^{- 3}})}^{- 3}$ $((2x^-6)/(7y^-3))^-3$ ?

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Answer 1 · 2016-01-09T10:55:37

$\frac{343 x^{18}}{8 y^{9}}$ $(343x^18)/(8y^9)$

Explanation:

A negative power $x^{- n}$ $x^(-n)$ can be seen as $x^{n^{- 1}}$ $x^(n^(-1))$ . We know that $x^{- 1} = \frac{1}{x}$ $x^(-1) = 1/x$ so $x^{n^{- 1}} = \frac{1}{x^{n}}$ $x^(n^(-1)) = 1/x^n$ , hence the following simplyfictations : $2 x^{- 6} = \frac{2}{x^{6}}$ $2x^-6 = 2/x^6$ and $7 y^{- 3} = \frac{7}{y^{3}}$ $7y^-3 = 7/y^3$

So ${(\frac{2 x^{- 6}}{7 y^{- 3}})}^{- 3} = {⎛ ⎝ \frac{\frac{2}{x^{6}}}{\frac{7}{y^{3}}} ⎞ ⎠}^{- 3} = {((\frac{2}{x^{6}}) \cdot (\frac{y^{3}}{7}))}^{- 3}$ $((2x^-6)/(7y^-3))^(-3) = ((2/x^6)/(7/y^3))^(-3) = ((2/x^6)*(y^3/7))^(-3)$ .

We do what we did before : ${((\frac{2}{x^{6}}) \cdot (\frac{y^{3}}{7}))}^{- 3} = {(\frac{x^{6}}{2} \cdot \frac{7}{y^{3}})}^{3} = \frac{343 x^{18}}{8 y^{9}}$ $((2/x^6)*(y^3/7))^(-3) = (x^6/2*7/y^3)^3 = (343x^18)/(8y^9)$

Answer 2 · 2016-01-09T16:18:55

A slightly different way or writing the working out!

$\frac{343 x^{18}}{8 y^{9}}$ $(343x^18)/(8y^9)$

Explanation:

Given: ${(\frac{2 x^{- 6}}{7 y^{- 3}})}^{- 3}$ $( (2x^(-6))/(7y^(-3)))^(-3)$

First consider inside the brackets:

Known that $. . 2 x^{- 6} \to \frac{2}{x^{6}} and that \frac{1}{7 y^{- 3}} \to \frac{y^{3}}{7}$ $color(white)(..)2x^(-6) -> 2/(x^6)" and that " 1/(7y^(-3)) ->y^3/7$

So we have $\frac{2}{x^{6}} \times \frac{y^{3}}{7} = \frac{2 y^{3}}{7 x^{6}}$ $2/x^6xxy^3/7= (2y^3)/(7x^6)$

Now put this back into the brackets giving

${(\frac{2 y^{3}}{7 x^{6}})}^{- 3} = {(\frac{7 x^{6}}{2 y^{3}})}^{3}$ $((2y^3)/(7x^6))^(-3) = ((7x^6)/(2y^3))^3$

$= \frac{343 x^{18}}{8 y^{9}}$ $=(343x^18)/(8y^9)$

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How do you simplify ${(\frac{2 x^{- 6}}{7 y^{- 3}})}^{- 3}$ $((2x^-6)/(7y^-3))^-3$ ?

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