MCQ
$2 \frac{\sqrt[3]{3}}{\sqrt[3]{25}}$ when written with a rational denominator is
- A$\frac{\sqrt[3]{15}}{5}$
- ✓$\frac{2}{5} \sqrt[3]{15}$
- C$2 \sqrt[3]{15}$
- Dnone of these
✓
Answer
Correct option: B.
$\frac{2}{5} \sqrt[3]{15}$
(b)
We have,
$2 \frac{\sqrt[3]{3}}{\sqrt[3]{25}}=2 \frac{\sqrt[3]{3}}{\sqrt[3]{5^2}}=2 \frac{\sqrt[3]{3}}{\sqrt[3]{5^2}} \times \frac{\sqrt[3]{5}}{\sqrt[3]{5}} \quad\left[\because \sqrt[3]{5}\right.$ is a rationalising factor of $\left.\sqrt[3]{5^2}\right]$
$=2 \frac{\sqrt[3]{3 \times 5}}{\sqrt[3]{5^3}}=2 \frac{\sqrt[3]{15}}{5} \quad\left[\because \sqrt[n]{a^n}=a\right]$
We have,
$2 \frac{\sqrt[3]{3}}{\sqrt[3]{25}}=2 \frac{\sqrt[3]{3}}{\sqrt[3]{5^2}}=2 \frac{\sqrt[3]{3}}{\sqrt[3]{5^2}} \times \frac{\sqrt[3]{5}}{\sqrt[3]{5}} \quad\left[\because \sqrt[3]{5}\right.$ is a rationalising factor of $\left.\sqrt[3]{5^2}\right]$
$=2 \frac{\sqrt[3]{3 \times 5}}{\sqrt[3]{5^3}}=2 \frac{\sqrt[3]{15}}{5} \quad\left[\because \sqrt[n]{a^n}=a\right]$
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