Respuesta :

Answer:

The mistake is the -3 exponent was not applied correctly to the coefficients (the 2 and the 4 in the original question) see image.

Step-by-step explanation:

The exponent rules being used here are applying an exponent to ALL the factors in the numerator (top part) and the denominator (bottom part). It is easier to simplify first. Then the -3 exponent goes on all the factors (the x, the y, and the 2) "Fix" negative exponents by pushing those terms across the fraction bar. Finally, 2^3 is 8. Nothing else can be simplified. See image.

Ver imagen lpina68

[tex]\left( \cfrac{2x^2}{4y^3} \right)^{-3}\implies \left( \cfrac{4y^3}{2x^2} \right)^{3}\implies \left( \cfrac{2y^3}{x^2} \right)^{3}\implies \cfrac{2^{1\cdot 3}y^{3\cdot 3}}{x^{2\cdot 3}}\implies \cfrac{2^3y^9}{x^6}\implies \cfrac{8y^9}{x^6}~\textit{\large\checkmark} \\\\[-0.35em] ~\dotfill\\\\ \left( \cfrac{2x^2}{4y^3} \right)^{-3}\implies \underset{\bigotimes}{\left( \cfrac{2y^3}{4x^2} \right)^{3}}\implies \left( \cfrac{y^3}{2x^2} \right)^{3}\implies \underset{\bigotimes}{\cfrac{y^9}{2x^6}}[/tex]

not carrying the constant with the term.

no distributing the exponent to the constant.

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