1. Problem statement: Prove that $\sqrt[3]{3}$ is irrational. 2. Formula and concept: A number is irrational if it cannot be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers with no common factors other than 1, and $q \neq 0$. 3. Assume the opposite (for contradiction): Suppose $\sqrt[3]{3}$ is rational. Then it can be written as $\frac{p}{q}$ in lowest terms, where $p$ and $q$ are integers with no common factors. 4. Write the equation: $$\sqrt[3]{3} = \frac{p}{q}$$ Cubing both sides gives: $$3 = \frac{p^3}{q^3}$$ 5. Multiply both sides by $q^3$: $$3q^3 = p^3$$ 6. This means $p^3$ is divisible by 3, so $p$ must be divisible by 3 (because 3 is prime). 7. Let $p = 3k$ for some integer $k$. Substitute back: $$3q^3 = (3k)^3 = 27k^3$$ 8. Simplify: $$3q^3 = 27k^3 \implies q^3 = 9k^3$$ 9. This means $q^3$ is divisible by 3, so $q$ is divisible by 3. 10. But this contradicts the assumption that $p$ and $q$ have no common factors (both divisible by 3). 11. Therefore, our initial assumption is wrong, and $\sqrt[3]{3}$ is irrational. Is tarah se humne prove kiya ki $\sqrt[3]{3}$ irrational hai.