Show that : a 3 k × a 3 l × a 3 m a 3 k + l + m = 1

Here we need to show that a 3 k × a 3 l × a 3 m a 3 k + l + m = 1 Now we have ⇒ a 3 k × a 3 l × a 3 m a 3 k + l + m = a 3 k × a 3 l × a 3 m a 3 k + l + m [Since a m n = a m n ] ⇒ a 3 k × a 3 l × a 3 m a 3 k + l + m = a 3 k + 3 l + 3 m a 3 k + l + m [Apply the law x a × x b × x c = x a + b + c ] ⇒ a 3 k + l + m a 3 k + l + m = 1 Hence proved the result.

Show that : a 2 k × a 2 l × a 2 m a k + l + m 2 = 1

Here we need to show that a 2 k × a 2 l × a 2 m a k + l + m 2 = 1 Now we have ⇒ a 2 k × a 2 l × a 2 m a k + l + m 2 = a 2 k × a 2 l × a 2 m a 2 k + l + m [Since a m n = a m n ] ⇒ a 2 k × a 2 l × a 2 m a 2 k + l + m = a 2 k + 2 l + 2 m a 2 k + l + m [Apply the law x a × x b × x c = x a + b + c ] ⇒ a 2 k + l + m a 2 k + l + m = 1 Hence proved the result.

E M B I B E

ICSE Skills In Mathematics>Algebra>Chapter 14 - Exponents>EXERCISE 14

V K Raman and S N Sharma Solutions for Exercise 1: EXERCISE 14

Author:V K Raman & S N Sharma

V K Raman Mathematics Solutions for Exercise - V K Raman and S N Sharma Solutions for Exercise 1: EXERCISE 14

Attempt the practice questions from Exercise 1: EXERCISE 14 with hints and solutions to strengthen your understanding. ICSE Skills In Mathematics solutions are prepared by Experienced Embibe Experts.