Question

$y = \frac{(190)^{3} {68^{2} } \times ( \frac{34}{95}^{ \frac{19}{3} }$
plz help me solve this question. ​

Answer 1

Answer:

  $\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$

Step-by-step explanation:

The applicable rules of exponents are ...

  (ab)^c = (a^c)(b^c)

  (a^b)/(a^c) = a^(b-c)

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  $\dfrac{190^3}{68^2}\times\dfrac{34}{95^{\frac{19}{3}}}=\dfrac{(2\cdot 95)^3}{(2\cdot 34)^2}\cdot\dfrac{34}{95^6\cdot 95^{\frac{1}{3}}}=2^{3-2}95^{3-6-\frac{1}{3}}34^{1-2}\\\\=2\cdot 95^{-3\frac{1}{3}}\cdot 34^{-1}=2\cdot 95^{-4+\frac{2}{3}}\cdot 34^{-1}\\\\=\dfrac{2\sqrt[3]{95^2}}{95^4\cdot 34}=\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}\\\\=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$