Is the fifth power of a number divisible by the cube of that number? Is the product of two cube numbers a cube number? Is q^k both a fourth power and a sixth power (q prime)?

You're on a roll — Step 1: Solve Question 4. c) किसी संख्या की पाँचवी घात उस संख्या के घन से विभाज्य होती है। (Is the fifth power of a number divisible by the cube of that number?) Let the number be x. Its fifth power is x^5 and its cube is x^3. To check divisibility, we look at the ratio: (x^5)/(x^3) = x^5-3 = x^2 Since x^2 is an integer for any integer x, the fifth power of a number is always divisible by the cube of that number (provided x ≠ 0). This statement is हमेशा सत्य है (always true). d) दो घन संख्याओं का गुणनफल एक घन संख्या होती है। (Is the product of two cube numbers a cube number?) Let the two cube numbers be a^3 and b^3. Their product is a^3 × b^3. Using the property of exponents, (ab)^3 = a^3 b^3. So, a^3 × b^3 = (ab)^3. Since (ab)^3 is a cube number, the product of two cube numbers is always a cube number. This statement is हमेशा सत्य है (always true). e) q^k एक चतुर्थ घात और एक छठी घात दोनों है (q एक अभाज्य संख्या है)। (q^k is both a fourth power and a sixth power (q is a prime number).) If q^k is a fourth power, then k must be a multiple of 4. So, k = 4m for some integer m. If q^k is a sixth power, then k must be a multiple of 6. So, k = 6n for some integer n. For k to be both a multiple of 4 and a multiple of 6, k must be a multiple of the least common multiple (LCM) of 4 and 6. LCM(4, 6) = 12. So, k must be a multiple of 12. For example, if k=12, then q^12 = (q^3)^4 (a fourth power) and q^12 = (q^2)^6 (a sixth power). Since such a k exists (e.g., k=12), this statement is कभी-कभी सत्य है (sometimes true). Send me the next one 📸