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For Competitive Exams

Simplify:
$\dfrac{\sqrt[3]{729}-\sqrt[3]{27}+\sqrt[2]{16}}{\sqrt[3]{512}+\sqrt[3]{343}-\sqrt[4]{256}}=$

$\dfrac{11}{10}$
$\dfrac{10}{11}$
$\dfrac{9}{10}$
$\dfrac{12}{11}$
Explanation:

$\dfrac{9-3+4}{8+7-4}=\dfrac{10}{11}$

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