Question:
Mark (✓) against the correct answer
$\sqrt[3]{\frac{-343}{729}}=?$
(a) $\frac{7}{9}$
(b) $\frac{-7}{9}$
(C) $\frac{-9}{7}$
(d) $\frac{9}{7}$
Solution:
(b) $\frac{-7}{9}$
By prime factorisation:
$\sqrt[3]{\frac{-343}{729}}=\frac{\sqrt[3]{-343}}{\sqrt[3]{729}}=\frac{\sqrt[3]{(-7) \times(-7) \times(-7)}}{\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3}}=\frac{\sqrt[3]{(-7)^{3}}}{\sqrt[3]{(3)^{3} \times(3)^{3}}}$
$\sqrt[3]{\frac{-343}{729}}=\frac{\sqrt[3]{(-7)^{3}}}{\sqrt[3]{(9)^{3}}}=\frac{-7}{9}$
$\therefore \sqrt[3]{\frac{-343}{729}}=\frac{-7}{9}$
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