MCQ
The product $\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}$ is equal to
- A$\sqrt[12]{2}$
- ✓$2$
- C$\sqrt{2}$
- D$\sqrt[12]{32}$
✓
Answer
Correct option: B.
$2$
$\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}$
$=\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{(2)^5}$
$=(2)^{\frac{1}{3}} \cdot(2)^{\frac{1}{4}} \cdot(2)^{\frac{5}{12}}$
$=(2)^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=(2)^{\frac{4+3+5}{12}}$
$=(2)^{\frac{12}{12}}$
$=2$
$=\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{(2)^5}$
$=(2)^{\frac{1}{3}} \cdot(2)^{\frac{1}{4}} \cdot(2)^{\frac{5}{12}}$
$=(2)^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=(2)^{\frac{4+3+5}{12}}$
$=(2)^{\frac{12}{12}}$
$=2$
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