Multiply x - 1 + - 3 2 and x - 1 - - 3 2 .

Let A = x - 1 + - 3 2 x - 1 - - 3 2 We know that - 1 = i Therefore, 1 + - 3 2 = 1 + i 3 2 , and 1 - - 3 2 = 1 - i 3 2 Thus, A = x - 1 + i 3 2 x - 1 - i 3 2 = x + - 1 - i 3 2 x + - 1 + i 3 2 = x + ω x + ω 2 , where ω , and ω 2 are cube roots of unity. = x 2 + ω 2 x + ω x + ω 3 = x 2 + ω 2 x + ω x + 1 ∵ ω 3 = 1 = x 2 + ω 2 + ω x + 1 = x 2 - x + 1 ∵ 1 + ω + ω 2 = 0 ⇒ ω + ω 2 = - 1 Hence, x - 1 + - 3 2 x - 1 - - 3 2 = x 2 - x + 1

Express with rational denominator: 1 3 - - 2

Given, 1 3 - - 2 = 1 3 - i 2 ∵ - 1 = i Multiplying and dividing the numerator and denominator with the conjugate of denominator, we have 1 3 - - 2 = 1 3 - i 2 × 3 + i 2 3 + i 2 Using, a + i b a - i b = a 2 - i b 2 = 3 + i 2 3 2 - i 2 2 = 3 + i 2 9 - 2 i 2 = 3 + i 2 9 - 2 - 1 ∵ i 2 = - 1 = 3 + i 2 11 Hence, 1 3 - - 2 = ( 3 + i 2 ) 11

Express with rational denominator: 3 - 2 + 2 ( - 5 ) 3 - 2 - 2 - 2

Given, 3 - 2 + 2 ( - 5 ) 3 - 2 - 2 - 2 We know that - 1 = i Therefore, 3 - 2 + 2 ( - 5 ) 3 - 2 - 2 - 2 = 3 2 i - 10 3 2 i - 2 2 i = 3 2 i - 10 2 i Now, multiplying, and dividing the numerator, and denominator with the conjugate of denominator, we have = 3 2 i - 10 2 i × - 2 i - 2 i = 3 2 i - 10 - 2 i - 2 i 2 = - 6 i 2 + 10 2 i - 2 i 2 = 6 + 10 2 i 2 ∵ i 2 = - 1 = 3 + 5 2 i Hence, 3 - 2 + 2 ( - 5 ) 3 - 2 - 2 - 2 = 3 + 5 2 i

Express with rational denominator: a + x i a - x i + a - x i a + x i

Given, a + x i a - x i + a - x i a + x i Taking L.C.M of the denominators, we have a + x i a - x i + a - x i a + x i = a + x i 2 + a - x i 2 a - x i a + x i = a 2 + i 2 x 2 + 2 a i x + a 2 + x 2 i 2 - 2 a i x a 2 - x i 2 [using the identities, a + b 2 = a 2 + 2 a b + b 2 , a + b a - b = a 2 - b 2 ] = 2 a 2 + 2 i 2 x 2 a 2 - i 2 x 2 = 2 a 2 - 2 x 2 a 2 + x 2 ∵ i 2 = - 1 Hence, a + x i a - x i + a - x i a + x i = 2 a 2 - 2 x 2 a 2 + x 2

Express with rational denominator: ( x + i ) 2 x - i + ( x - - 1 ) 2 ( x + i )

Given, ( x + i ) 2 x - i + ( x - - 1 ) 2 ( x + i ) We know that - 1 = i ⇒ ( x + i ) 2 x - i + ( x - - 1 ) 2 ( x + i ) = ( x + i ) 2 x - i + x - i 2 x + i Taking L.C.M of the denominators, we have ( x + i ) 2 x - i + ( x - - 1 ) 2 ( x + i ) = x + i 3 + x - i 3 x - i x + i = x 3 + i 3 + 3 x 2 i + 3 x i 2 + x 3 - i 3 - 3 x 2 i + 3 x i 2 x 2 - i 2 [using the identities, a + b 3 = a 3 + b 3 + 3 a 2 b + 3 a b 2 , a - b 3 = a 3 - b 3 - 3 a 2 b + 3 a b 2 , and a + b a - b = a 2 - b 2 ] = 2 x 3 + 6 x i 2 x 2 + 1 = 2 x 3 - 6 x x 2 + 1 ∵ i 2 = - 1 = 2 x x 2 - 3 x 2 + 1

Express with rational denominator: ( a + i ) 3 - ( a - i ) 3 ( a + i ) 2 - ( a - i ) 2

Given, ( a + i ) 3 - ( a - i ) 3 ( a + i ) 2 - ( a - i ) 2 We know that x 3 - y 3 = x - y x 2 + x y + y 2 , and x 2 - y 2 = x - y x + y Therefore, ( a + i ) 3 - ( a - i ) 3 ( a + i ) 2 - ( a - i ) 2 = a + i - a + i a + i 2 + a + i a - i + a - i 2 a + i - a + i a + i + a - i = 2 i a 2 + i 2 + 2 a i + a 2 - i 2 + a 2 + i 2 - 2 a i 2 i × 2 a [using the identities, x + y 2 = x 2 + 2 x y + y 2 , x - y x + y = x 2 - y 2 ] = 3 a 2 + i 2 2 a = 3 a 2 - 1 2 a ∵ i 2 = - 1 Hence, ( a + i ) 3 - ( a - i ) 3 ( a + i ) 2 - ( a - i ) 2 = 3 a 2 - 1 2 a

Find the value of - - 1 4 n + 3 , when n is a positive integer.

We have, - - 1 4 n + 3 We know that - 1 = i iota . Therefore, we get - - 1 4 n + 3 = - i 4 n + 3 = - i 4 n × - i 3 = - i 4 n × - - i ; ∵ i 3 = - i = 1 × i = i ∵ i 4 = 1 and i 2 = - 1 )