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Find the cube root of each of the following rational numbers.
(i) $ \frac{-125}{729} $(ii) $ \frac{10648}{12167} $
(iii) $ \frac{-19683}{24389} $
(iv) $ \frac{686}{-3456} $
(v) $ \frac{-39304}{-42875} $
To find:
We have to find the cube roots of the given rational numbers.
Solution:
(i) $\sqrt[3]{\frac{-125}{729}} =\frac{\sqrt[3]{-125}}{\sqrt[3]{729}}$
$=\frac{-\sqrt[3]{125}}{\sqrt[3]{729}}$
$=\frac{-\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3}}$
$=\frac{-\sqrt[3]{5^{3}}}{\sqrt[3]{3^{3} \times 3^{3}}}$
$=\frac{-5}{3 \times 3}$
$=\frac{-5}{9}$
(ii) $\sqrt[3]{\frac{10648}{12167}}=\frac{\sqrt[3]{10648}}{\sqrt[3]{12167}}$
$=\frac{\sqrt[3]{2 \times 2 \times 2 \times 11 \times 11 \times 11}}{\sqrt[3]{23 \times 23 \times 23}}$
$=\frac{\sqrt[3]{2^{3} \times 11^{3}}}{\sqrt[3]{23^{3}}}$
$=\frac{2 \times 11}{23}$
$=\frac{22}{23}$
(iii) $\sqrt[3]{\frac{-19683}{24389}}=\frac{\sqrt[3]{-19683}}{\sqrt[3]{24389}}$
$=\frac{-\sqrt[3]{19683}}{\sqrt[3]{24389}}$
$=\frac{-\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}}{\sqrt[3]{29 \times 29 \times 29}}$
$=\frac{-\sqrt[3]{3^{3} \times 3^{3} \times 3^{3}}}{\sqrt[3]{29^{3}}}$
$=\frac{-(3 \times 3 \times 3)}{29}$
$=\frac{-27}{29}$
(iv) $\sqrt[3]{\frac{686}{-3456}}=\sqrt[3]{\frac{2 \times 343}{-2 \times 1728}}$
$=\sqrt[3]{\frac{343}{-1728}}$
$=\frac{\sqrt[3]{343}}{\sqrt[3]{-1728}}$
$=\frac{\sqrt[3]{343}}{-\sqrt[3]{1728}}$
$=\frac{\sqrt[3]{7 \times 7 \times 7}}{-\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}}$
$=\frac{\sqrt[3]{7^{3}}}{-\sqrt[3]{2^{3} \times 2^{3} \times 3^{3}}}$
$=\frac{7}{-(2 \times 2 \times 3)}$
$=\frac{-7}{12}$
(v) $\sqrt[3]{\frac{-39304}{-42875}}=\sqrt[3]{\frac{39304}{42875}}$
$=\frac{\sqrt[3]{39304}}{\sqrt[3]{42875}}$
$=\frac{\sqrt[3]{2 \times 2 \times 2 \times 17 \times 17 \times 17}}{\sqrt[3]{5 \times 5 \times 5 \times 7 \times 7 \times 7}}$
$=\frac{\sqrt[3]{2^{3} \times 17^{3}}}{\sqrt[3]{5^{3} \times 7^{3}}}$
$=\frac{2 \times 17}{5 \times 7}$
$=\frac{34}{35}$