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1 vote
1 vote
What did they do wrong?
give details please....

What did they do wrong? give details please....-example-1
User Toadjaune
by
2.3k points

2 Answers

12 votes
12 votes


\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}}

not carrying the constant with the term.

no distributing the exponent to the constant.

User Yaqub Ahmad
by
3.0k points
14 votes
14 votes

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.

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.

What did they do wrong? give details please....-example-1
User Joseph Leedy
by
3.1k points