Question

Mark (✓) against the correct answer
$\sqrt[3]{\frac{-343}{729}}=?$
(a) $\frac{7}{9}$
(b) $\frac{-7}{9}$
(c) $\frac{-9}{7}$
(d) $\frac{9}{7}$

Answer 1

(b) $\frac{-7}{9}$
By prime factorisation:
$\sqrt[3]{\frac{-343}{729}}$ = $\frac{\sqrt[3]{-343}}{\sqrt[3]{729}}=\frac{\sqrt[3]{\left(-7\right)\times \left(-7\right)\times \left(-7\right)}}{\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3}}=\frac{\sqrt[3]{{\left(-7\right)}^3}}{\sqrt[3]{{\left(3\right)}^3\times {\left(3\right)}^3}}$
$\sqrt[3]{\frac{-343}{729}}$ = $\frac{\sqrt[3]{{\left(-7\right)}^3}}{\sqrt[3]{{\left(9\right)}^3}}$ = $\frac{-7}{9}$

∴ $\sqrt[3]{\frac{-343}{729}}$ = $\frac{-7}{9}$