Question

by what number should (3/-2)^-3 be divided to get (2/3)^2 ?

Answer 1

Let $x$ be the number that should divide $(\frac{3}{-2} )^{-3}$ to get $(\frac{2}{3} )^2$. Then

$\frac{(\frac{3}{-2} )^{-3}}{x}=(\frac{2}{3} )^2 \\ \frac{(\frac{-2}{3} )^{3}}{x}=(\frac{2}{3} )^2 \\ \frac{-8}{27x}=\frac{4}{9} \\ \frac{4}{9} =\frac{-8}{27x}\\ x=-\frac{72}{108} \\ x=-\frac{2}{3}$

The the number that should divide $(\frac{3}{-2} )^{-3}$ is $-\frac{2}{3}$.

Answer 2

$\textbf {Hey there, here is the solution.}$

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Let the number be x

(3/-2)⁻³ ÷ x = (2/3) ²

Convert the exponent to positive exponents:

(-2/3)³ ÷  = (2/3)²

Evaluate the exponents

-8/27 ÷ x = 4/9

Convert the divide into multiply:

-8/27 x 1/x = 4/9

Combine into single fraction:

-8/27x = 4/9

Cross multiply:

4 x 27x = -8 x 9

Evaluate each side:

108x = -72

Divide both sides by 108:

x = -72/108

simplify:

x = -2/3

Answer: The number is -2/3

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⭐$\textbf {Cheers}$⭐