Mark (✓) against the correct answer

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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