If x 2 + 1 x 2 = 23 , then find the value of x + 1 x and x 3 + 1 x 3

Given x 2 + 1 x 2 = 23 We know that a + b 2 = a 2 + 2 a b + b 2 x + 1 x 2 = x 2 + 2 x × 1 x + 1 x 2 x + 1 x 2 = x 2 + 2 + 1 x 2 = x 2 + 1 x 2 + 2 x + 1 x 2 = 23 + 2 = 25 x + 1 x = 25 = ± 5 . . . 1 We know that a + b 3 = a 3 + b 3 + 3 a b a + b x 3 + 1 x 3 = x + 1 x 3 - 3 x × 1 x x + 1 x = 5 3 - 3 ( 5 ) = 125 - 15 = 110 [when x + 1 x = 5 ] Also, = - 5 3 - 3 × - 5 = - 125 + 15 = - 110 [when x + 1 x = - 5 ] Hence, x 3 + 1 x 3 = ± 110 Therefore, x + 1 x = ± 5 and x 3 + 1 x 3 = ± 110 .

If x 2 + 1 x 2 = 23 , then find the value of x + 1 x and x 3 + 1 x 3

Given x 2 + 1 x 2 = 23 We know that a + b 2 = a 2 + 2 a b + b 2 x + 1 x 2 = x 2 + 2 x × 1 x + 1 x 2 x + 1 x 2 = x 2 + 2 + 1 x 2 = x 2 + 1 x 2 + 2 x + 1 x 2 = 23 + 2 = 25 x + 1 x = 25 = ± 5 . . . 1 We know that a + b 3 = a 3 + b 3 + 3 a b a + b x 3 + 1 x 3 = x + 1 x 3 - 3 x × 1 x x + 1 x = 5 3 - 3 ( 5 ) = 125 - 15 = 110 [when x + 1 x = 5 ] Also, = - 5 3 - 3 × - 5 = - 125 + 15 = - 110 [when x + 1 x = - 5 ] Hence, x 3 + 1 x 3 = ± 110 Therefore, x + 1 x = ± 5 and x 3 + 1 x 3 = ± 110 .

Prove: a + 1 a 2 + a - 1 a 2 = 2 a 2 + 1 a 2 2

To prove a + 1 a 2 + a - 1 a 2 = 2 a 2 + 1 a 2 2 We know that ( a + b ) 2 = a 2 + 2 a b + b 2 and ( a - b ) 2 = a 2 - 2 a b + b 2 . So, a + 1 a 2 = a 2 + 1 a 2 + 2 a × 1 a = 2 + a 2 + 1 a 2 . . . . . ( i ) a - 1 a 2 = a 2 + 1 a 2 - 2 a × 1 a = a 2 + 1 a 2 - 2 . . . . ( ii ) Add both the equations 2 + a 2 + 1 a 2 + a 2 + 1 a 2 - 2 2 a 2 + 2 a 2 2 a 2 + 1 a 2 ∴ a + 1 a 2 + a - 1 a 2 = 2 a 2 + 2 × 1 a 2 = 2 a 2 + 1 a 2 = RHS Hence proved.

Prove: a + 1 a 2 - a - 1 a 2 = 4

To prove a + 1 a 2 - a - 1 a 2 = 4 . We know that x + y = x 2 + 2 x y + y 2 and x - y = x 2 - 2 x y + y 2 . LHS : a + 1 a 2 - a - 1 a 2 = a 2 + 1 a 2 + 2 a × 1 a - a 2 + 1 a 2 - 2 a × 1 a = a 2 + 1 a 2 + 2 - a 2 - 1 a 2 + 2 = 4 = RHS Hence proved.