Find the factor which will rationalise 5 3 - 3 4 .

Given, 5 3 - 3 4 Put x = 5 1 3 ⇒ x 12 = 5 4 and y = 3 1 4 ⇒ y 12 = 3 3 , then x 12 and y 12 both are rational. Now, x 12 - y 12 = x 6 - y 6 x 6 + y 6 = x 3 - y 3 x 3 + y 3 x 6 + y 6 = x - y x 2 + x y + y 2 x 3 + y 3 x 6 + y 6 = x - y x 2 + x y + y 2 x 9 + x 3 y 6 + y 3 x 6 + y 9 = ( x - y ) x 11 + x 10 y + x 9 y 2 + x 8 y 3 + x 7 y 4 + ⋯ + y 11 Therefore, the rationalising factor is x 11 + x 10 y + x 9 y 2 + … … … + y 11 = 5 11 3 + 5 10 3 · 3 1 4 + 5 9 3 · 3 1 2 + … … … + 3 11 4

Find the factor which will rationalise 5 3 - 3 4 .

Given, 5 3 - 3 4 Put x = 5 1 3 ⇒ x 12 = 5 4 and y = 3 1 4 ⇒ y 12 = 3 3 , then x 12 and y 12 both are rational. Now, x 12 - y 12 = x 6 - y 6 x 6 + y 6 = x 3 - y 3 x 3 + y 3 x 6 + y 6 = x - y x 2 + x y + y 2 x 3 + y 3 x 6 + y 6 = x - y x 2 + x y + y 2 x 9 + x 3 y 6 + y 3 x 6 + y 9 = ( x - y ) x 11 + x 10 y + x 9 y 2 + x 8 y 3 + x 7 y 4 + ⋯ + y 11 Therefore, the rationalising factor is x 11 + x 10 y + x 9 y 2 + … … … + y 11 = 5 11 3 + 5 10 3 · 3 1 4 + 5 9 3 · 3 1 2 + … … … + 3 11 4

Express with rational denominator 3 3 - 1 3 3 + 1

Given, 3 3 - 1 3 3 + 1 In order to rationalise 3 1 3 - 1 3 1 3 + 1 , we will first obtain the rationalising factor of 3 1 3 + 1 Put x = 3 1 3 ⇒ x 3 = 1 and y = 1 ⇒ y 3 = 1 , then we have x 3 + y 3 = ( x + y ) x 2 - x y + y 2 Therefore, the rationalising factor is x 2 - x y + y 2 = 3 2 3 - 3 1 3 + 1 . . . i Now multiplying and dividing the given expression by i , we have 3 1 3 - 1 3 2 3 - 3 1 3 + 1 3 1 3 + 1 3 2 3 - 3 1 3 + 1 = 3 - 2 × 3 2 3 + 2 × 3 1 3 - 1 3 1 3 3 + ( 1 ) 3 = 2 - 2 × 3 2 3 + 2 × 3 1 3 4 = 1 - 3 2 3 + 3 1 3 2

Express 9 6 - 8 6 9 6 + 8 6 with rational denominator.

Given, 9 6 - 8 6 9 6 + 8 6 Now, Put x = 9 1 6 ⇒ x 6 = 9 and y = 8 1 6 ⇒ y 6 = 8 Then, x 6 - y 6 = x 3 + y 3 x 3 - y 3 = x + y x 2 - x y + y 2 x 3 - y 3 = x + y x 5 - x 4 y + x 3 y 2 - x 2 y 3 + x y 4 - y 5 Therefore, the rational factor is x 5 - x 4 y + x 3 y 2 - x 2 y 3 + x y 4 - y 5 = 9 5 6 - 9 4 6 · 8 1 6 + 9 3 6 · 8 2 6 - … - 8 5 6 Multiplying the given expression by this rationalizing factor above and below and simplifying, we get 9 1 6 - 8 1 6 9 1 6 + 8 1 6 × 9 5 6 - 9 4 6 · 8 1 6 + 9 3 6 · 8 2 6 - … - 8 5 6 9 5 6 - 9 4 6 · 8 1 6 + 9 3 6 · 8 2 6 - … - 8 5 6 = 17 - 2 × 3 5 3 × 2 1 3 + 2 · 3 4 3 · 2 1 - 2 · 3 1 · 2 3 2 + 2 · 3 2 3 · 2 2 - 2 · 3 1 3 · 2 5 2 9 - 8 = 17 - 3 5 3 · 2 4 3 + 3 4 3 · 2 2 - 3 · 2 5 2 + 2 3 · 3 2 3 - 3 1 3 · 2 7 2

Express with rational denominator 2 - 3 3 3 3 + 2

Given, 2 - 3 3 3 3 + 2 Put x = 3 3 ⇒ x 6 = 3 2 and y = 2 ⇒ y 6 = 2 3 . Then, we have x 6 - y 6 = x 3 - y 3 x 3 + y 3 = ( x + y ) x 2 - x y + y 2 x 3 - y 3 = ( x + y ) x 5 - x 4 y + x 3 y 2 - x 2 y 3 + x y 4 - y 5 Then the rationalising factor is x 5 - x 4 y + x 3 y 2 - x 2 y 3 + x y 4 - y 5 = 3 5 3 - 3 4 3 · 2 1 2 + 3 1 · 2 1 - 3 2 3 · 2 3 2 + 3 1 3 · 2 2 - 2 5 2 Multiplying and dividing the given fraction by this rationalising factor and simplifying, we get 2 1 2 - 3 1 3 3 1 3 + 2 1 2 × 3 5 3 - 3 4 3 · 2 1 2 + 3 1 · 2 1 - 3 2 3 · 2 3 2 + 3 1 3 · 2 2 - 2 5 2 3 5 3 - 3 4 3 · 2 1 2 + 3 1 · 2 1 - 3 2 3 · 2 3 2 + 3 1 3 · 2 2 - 2 5 2 = 3 6 3 · 2 1 2 - 3 5 3 · 2 1 + 3 4 3 · 2 3 2 - 3 1 · 2 2 + 3 2 3 · 2 5 2 - 2 3 · 3 1 3 9 - 8 = 3 2 · 2 1 2 - 3 5 3 · 2 + 3 4 3 · 2 3 2 - 3 · 2 2 + 3 2 3 · 2 5 2 - 3 1 3 · 2 3

Express with rational denominator 3 3 3 + 9 6

Given, 3 3 3 + 9 6 First we will find the rationalising factor of denominator. Put x = 3 1 2 ⇒ x 6 = 3 3 and y = 9 1 6 ⇒ y 6 = 9 , then we have x 6 - y 6 = x 3 + y 3 x 3 - y 3 = ( x + y ) x 2 - x y + y 2 x 3 - y 3 = ( x + y ) x 5 - x 4 y + x 3 y 2 - x 2 y 3 + x y 4 - y 5 Thus, the rationalising factor is x 5 - x 4 y + x 3 y 2 - x 2 y 3 + x y 4 - y 5 3 5 2 - 3 2 · 9 1 6 + 3 3 2 · 9 1 3 - 3 1 · 9 1 2 + 3 1 2 · 9 2 3 - 9 5 6 Multiplying and dividing the given fraction by this rationalising factor and simplifying, we get 3 3 3 + 9 6 × 3 5 2 - 3 2 · 9 1 6 + 3 3 2 · 9 1 3 - 3 1 · 9 1 2 + 3 1 2 · 9 2 3 - 9 5 6 3 5 2 - 3 2 · 9 1 6 + 3 3 2 · 9 1 3 - 3 1 · 9 1 2 + 3 1 2 · 9 2 3 - 9 5 6 = 3 12 6 · 3 5 6 - 3 6 3 · 3 1 3 · 3 1 3 + 3 11 6 · 3 4 6 - 3 4 3 · 3 1 + 3 5 6 · 3 4 3 - 3 10 6 · 3 1 3 27 - 9 = 9 3 5 6 - 3 2 3 + 3 1 2 - 3 1 3 + 3 1 6 - 1 18 = 1 2 3 5 6 - 3 2 3 + 3 1 2 - 3 1 3 + 3 1 6 - 1

Express with rational denominator 8 + 4 3 8 - 4 3

Given, 8 + 4 3 8 - 4 3 For rationalising the denominator Put x = 8 1 2 , y = 4 1 3 Then, x 6 - y 6 = x 3 + y 3 x 3 - y 3 = ( x - y ) x 2 + x y + y 2 x 3 + y 3 = ( x - y ) x 5 + x 4 y + x 3 y 2 + x 2 y 3 + x y 4 + y 5 Therefore, rationalisation factor of denominator is x 5 + x 4 y + x 3 y 2 + x 2 y 3 + x y 4 + y 5 ⇒ 8 5 2 + 8 4 2 · 4 1 3 + 8 3 2 · 4 2 3 + 8 1 · 4 1 + 8 1 2 · 4 4 3 + 4 5 3 Multiplying by it the numerator and denominator of 8 + 4 3 8 - 4 3 , we have ( 8 + 4 3 ) 8 5 2 + 8 4 2 · 4 1 3 + … + 4 5 3 8 1 2 - 4 1 3 8 5 2 + 8 4 2 · 4 1 3 + … + 4 5 3 = 2 9 + 2 55 6 + 2 50 6 + 2 45 6 + 2 40 6 + 2 35 6 + 2 24 6 ( 8 ) 3 - ( 4 ) 2 = 16 2 5 + 2 31 6 + 2 26 6 + 2 21 6 + 2 16 6 + 2 11 6 + 1 496 = 2 5 + 2 31 6 + 2 26 6 + 2 21 6 + 2 16 6 + 2 11 6 + 1 31

Express with rational denominator 27 6 3 - 9 6

Given, 27 6 3 - 9 6 = ( 27 ) 1 6 3 - 3 1 3 ∵ 9 1 6 = 3 1 3 = ( 3 ) 3 6 3 - 3 1 3 ; ∵ 27 1 6 = 3 3 6 In order to rationalise ( 3 ) 1 2 3 - 3 1 3 , we will first obtain the rationalising factor of 3 - 3 1 3 Put x = 3 & y = 3 1 3 , then x 3 - y 3 = ( x - y ) x 2 + x y + y 2 Therefore, the rationalising factor is x 2 + x y + y 2 or 3 2 + 3 · 3 1 3 + 3 2 3 Now multiplying and dividing the given expression by 3 2 + 3 · 3 1 3 + 3 2 3 , we get = ( 3 ) 3 6 3 2 + 3 · 3 1 3 + 3 2 3 3 - 3 1 3 3 2 + 3 · 3 1 3 + 3 2 3 = 3 3 / 6 3 2 + 3 · 3 4 3 + 3 2 3 27 - 3 = 3 3 9 6 + 3 5 6 + 3 1 6 24 = 3 3 2 + 3 5 6 + 3 1 6 8

Find the square root of 16 - 2 20 - 2 28 + 2 35

Let ( 16 - 2 20 - 2 28 + 2 35 ) = x - y - z Squaring both sides, we get 16 - 2 20 - 2 28 + 2 35 = x + y + z - 2 x y - 2 x z + 2 y z Equating rational and irrational parts of both sides x + y + z = 16 . . . ( i ) x y = 20 . . . ( ii ) x z = 28 . . . ( iii ) y z = 35 . . . ( iv ) From ( ii ) , ( iii ) & ( iv ) , we get x 2 y 2 z 2 = 20 × 28 × 35 or x y z = ± 140 Then with the help of ( ii ) , ( iii ) & ( iv ) , we have z = ± 7 , y = ± 5 , x = ± 4 But x + y + z = 16 ∴ Taking + ve sign, x = 4 , y = 5 , z = 7 which satisfy this relation i . Hence, the required square root is ± ( 4 - 5 - 7 ) or ± ( 2 - 5 - 7 )

Find the square root of 24 + 4 15 - 4 21 - 2 35

Let 24 + 4 15 - 4 21 - 2 35 = x - y - z , where x , y , z > 0 Squaring both sides, we get 24 + 4 15 - 4 21 - 2 35 = ( x - y - z ) 2 ⇒ 24 + 4 15 - 4 21 - 2 35 = x + y + z - 2 x y - 2 x z + 2 y z Equating the rational and irrational parts of both sides x + y + z = 24 . . . ( i ) x y = 21 × 4 = 84 . . . ( ii ) y z = 15 × 4 = 60 . . . ( iii ) z x = 35 . . . ( iv ) From ( ii ) , ( iii ) & ( iv ) , we get x 2 y 2 z 2 = 84 × 60 × 35 or x y z = 420 . Thus, from ( ii ) , ( iii ) & ( iv ) , we get x = 7 , y = 12 , z = 5 which satisfy the relation ( i ) Hence, the required square root is ± ( 7 - 5 - 2 3 )