Find the value of x 2 + 1 x 2 and x 4 + 1 x 4 if x + 1 x = 3 .

Given, x + 1 x = 3 We need to find the value of x 2 + 1 x 2 . We know that, x + 1 x 2 = x 2 + 1 x 2 + 2 Substituting the values, ⇒ ( 3 ) 2 = x 2 + 1 x 2 + 2 ⇒ x 2 + 1 x 2 = 9 - 2 ⇒ x 2 + 1 x 2 = 7 Similarly, x 4 + 1 x 4 = x 2 + 1 x 2 2 - 2 Substituting the values, = ( 7 ) 2 - 2 = 47 Hence, x 2 + 1 x 2 = 7 and x 4 + 1 x 4 = 47

Find the value of x - 1 x and x 2 - 1 x 2 , if x + 1 x = 5

Given, x + 1 x = 5 We need to find the value of x - 1 x . We know that, x - 1 x 2 = x + 1 x 2 - 4 Substituting the values, ⇒ x - 1 x 2 = 5 2 - 4 ⇒ x - 1 x 2 = 5 - 4 ⇒ x - 1 x 2 = 1 ⇒ x - 1 x = ± 1 ⇒ x - 1 x = ± 1 We know that, x 2 - 1 x 2 = x - 1 x x + 1 x Substituting the values, ⇒ x 2 - 1 x 2 = ± 1 × 5 ⇒ x 2 - 1 x 2 = ± 5 Hence, x - 1 x = ± 1 and x 2 - 1 x 2 = ± 5

Find the value of x 3 - 1 x 3 , if x 2 + 1 x 2 = 18

Given: x 2 + 1 x 2 = 18 We know that, x - 1 x 2 = x 2 + 1 x 2 - 2 Substituting the known values, ⇒ x - 1 x 2 = 18 - 2 ⇒ x - 1 x 2 = 16 ⇒ x - 1 x = ± 4 . . . . . 1 Now take cube on both side of ( 1 ) , When x - 1 x = 4 , ⇒ x 3 - 1 x 3 - 3 × x × 1 x x - 1 x = 64 ⇒ x 3 - 1 x 3 = 64 + 12 = 76 When x - 1 x = - 4 , ⇒ x 3 - 1 x 3 - 3 × x × 1 x x - 1 x = 64 ⇒ x 3 - 1 x 3 = - 64 - 12 = - 76 Hence, the value of x 3 - 1 x 3 is ± 76 .

Find the value of x 3 + 1 125 x 3 , if x 2 + 1 25 x 2 = 8 3 5

Given: x 2 + 1 25 x 2 = 8 3 5 We know that, x + 1 x 2 = x 2 + 1 x 2 + 2 x 1 x So, x + 1 5 x 2 = ( x ) 2 + 1 5 x 2 + 2 ( x ) 1 5 x ⇒ x + 1 5 x 2 = x 2 + 1 25 x 2 + 2 5 ⇒ x + 1 5 x 2 = 8 3 5 + 2 5 ⇒ x + 1 5 x 2 = 9 ⇒ x + 1 5 x = ± 3 . . . . . 1 Taking the cube of LHS of ( 1 ) , ⇒ x + 1 5 x 3 = ( x ) 3 + 1 5 x 3 + 3 ( x ) 1 5 x x + 1 5 x ⇒ x + 1 5 x 3 = x 3 + 1 125 x 3 + 3 5 x + 1 5 x . . . . . 2 When x + 1 5 x = + 3 in ( 2 ) , ⇒ ( 3 ) 3 = x 3 + 1 125 x 3 + 3 5 ( 3 ) ⇒ x 3 + 1 125 x 3 = 27 - 9 5 ⇒ x 3 + 1 125 x 3 = 25 1 5 When x + 1 5 x = - 3 in 2 , ⇒ ( - 3 ) 3 = x 3 + 1 125 x 3 + 3 5 ( - 3 ) ⇒ - 27 = x 3 + 1 125 x 3 - 9 5 ⇒ x 3 + 1 125 x 3 = - 27 + 9 5 ⇒ x 3 + 1 125 x 3 = - 25 1 5 Hence, x 3 + 1 125 x 3 = ± 25 1 5

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ICSE Mathematics>Algebra>Chapter 3 - Expansions>Exercise

O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 2: Exercise

Author:O P Malhotra, S K Gupta & Anubhuti Gangal

O P Malhotra Mathematics Solutions for Exercise - O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 2: Exercise

Attempt the practice questions from Exercise 2: Exercise with hints and solutions to strengthen your understanding. ICSE Mathematics solutions are prepared by Experienced Embibe Experts.