Question 43

The value of $$(3^2)^3 \times 3^{-5} \times \sqrt[3]{216} \div \sqrt{9} \times \sqrt[3]{8}$$ is

Solution

First, $$(3^2)^3 \times 3^{-5} = 3^6 \times 3^{-5} = 3$$. Also $$\sqrt[3]{216} = 6$$, $$\sqrt{9} = 3$$ and $$\sqrt[3]{8} = 2$$. Working left to right, $$3 \times 6 \div 3 \times 2 = 12$$.

Was this solution helpful?

Get AI Help

Other Cracku’s Useful Resources:

JEE Question Bank JEE Mains Previous Papers JEE Advanced Previous Papers JEE Study Material PDF JEE Mains Mock Test JEE Advanced Mock Test JEE Daily Target

Free JEE Questions

JEE Physics Questions JEE Maths Questions JEE Chemistry Questions
banner

banner

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now
Ask AI