## Algebra

#### Algebra

1.  If p = 3 + 1 , the value of p4 + 1 is : p p4
1. 81
2. 27
3. 120
4. 119

1.  p = 3 + 1 (Given) p

 ∴ p - 1 = 3 p

On squaring both sides,
 ⇒ p - 1 2 = (3)2 = 9 p

 ⇒ p2 + 1 - 2 = 9 p2

 ⇒ p2 + 1 = 9 + 2 = 11 p2

On squaring again,
 ⇒ p2 + 1 2 = (11)2 = 121 p2

 ⇒ p4 + 1 + 2 = 121 p4

 ⇒ p4 + 1 = 121 - 2 = 119 p4

##### Correct Option: D

 p = 3 + 1 (Given) p

 ∴ p - 1 = 3 p

On squaring both sides,
 ⇒ p - 1 2 = (3)2 = 9 p

 ⇒ p2 + 1 - 2 = 9 p2

 ⇒ p2 + 1 = 9 + 2 = 11 p2

On squaring again,
 ⇒ p2 + 1 2 = (11)2 = 121 p2

 ⇒ p4 + 1 + 2 = 121 p4

 ⇒ p4 + 1 = 121 - 2 = 119 p4

1. If x = 1 + √2 + √3, then the value of (2x4 – 8x3 + 26x – 28) is
1. 2√2
2. 3√3
3. 5√5
4. 6√6

1. x = 1 + √2 + √3
⇒ x - 1 = √3 + √2
On squaring both sides ,
x2 – 2x + 1 = 3 + 2 + 2√6
⇒ x2 – 2x + 1 - 5 = 2√6
⇒ x2 – 2x - 4 = 2√6
On squaring again,
(x2 – 2x - 4)2 = (2√6)2
⇒ x4 + 4x2 + 16 – 4x3 + 16x - 8x2 = 24
⇒ x4 – 4x3 - 4x2 + 16x - 8x2 - 8 = 0
⇒ 2x4 – 8x3 - 8x2 + 32x - 16 = 0
∴ 2x4 – 8x3 - 5x2 + 32x - 16 + 3x2 - 6x - 12
= 0 + 3 (x2 – 2x – 4) = 3 × 2√6
Required answer = 6√6

##### Correct Option: D

x = 1 + √2 + √3
⇒ x - 1 = √3 + √2
On squaring both sides ,
x2 – 2x + 1 = 3 + 2 + 2√6
⇒ x2 – 2x + 1 - 5 = 2√6
⇒ x2 – 2x - 4 = 2√6
On squaring again,
(x2 – 2x - 4)2 = (2√6)2
⇒ x4 + 4x2 + 16 – 4x3 + 16x - 8x2 = 24
⇒ x4 – 4x3 - 4x2 + 16x - 8x2 - 8 = 0
⇒ 2x4 – 8x3 - 8x2 + 32x - 16 = 0
∴ 2x4 – 8x3 - 5x2 + 32x - 16 + 3x2 - 6x - 12
= 0 + 3 (x2 – 2x – 4) = 3 × 2√6
Required answer = 6√6

1.  If x + 1 = √3 the value of (x18 + x12 + x6 + 1) is x
1. 0
2. 1
3. 2
4. 3

1.  x + 1 = √3 x

On cubing both sides,
 ⇒ x + 1 3 = (3√3)3 x

 ⇒ x3 + 1 + 3x. 1 x + 1 = 3√3 x3 x x

 ⇒ x3 + 1 + 3 × √3 = 3√3 x3

 ⇒ x3 + 1 = 3√3 - 3√3 = 0 x3

⇒ x6 + 1 = 0
∴ x18 + x12 + x6 + 1 = x12( x6 + 1 ) + 1( x6 + 1 )
x18 + x12 + x6 + 1 = ( x6 + 1 )( x12 + 1 ) = 0

##### Correct Option: A

 x + 1 = √3 x

On cubing both sides,
 ⇒ x + 1 3 = (3√3)3 x

 ⇒ x3 + 1 + 3x. 1 x + 1 = 3√3 x3 x x

 ⇒ x3 + 1 + 3 × √3 = 3√3 x3

 ⇒ x3 + 1 = 3√3 - 3√3 = 0 x3

⇒ x6 + 1 = 0
∴ x18 + x12 + x6 + 1 = x12( x6 + 1 ) + 1( x6 + 1 )
x18 + x12 + x6 + 1 = ( x6 + 1 )( x12 + 1 ) = 0

1. If a = 299, b = 298, c = 297 then the value of 2a3 + 2b3 + 2c3 – 6abc is
1. 5154
2. 5267
3. 5364
4. 5456

1. 2a3 + 2b3 + 2c3 – 6abc
= 2 (a3 + b3 + c3 – 3abc)

 = 2 (a + b + c) × 1 {(a – b)2 + (b –c)2 + (c – a)2} 2

= (299 + 298 + 297) {(299 – 298)2 + (298 – 297)2 + (297 – 299)2}
= 894 × (1 + 1 + 4)
= 894 × 6 = 5364

##### Correct Option: C

2a3 + 2b3 + 2c3 – 6abc
= 2 (a3 + b3 + c3 – 3abc)

 = 2 (a + b + c) × 1 {(a – b)2 + (b –c)2 + (c – a)2} 2

= (299 + 298 + 297) {(299 – 298)2 + (298 – 297)2 + (297 – 299)2}
= 894 × (1 + 1 + 4)
= 894 × 6 = 5364

1. If a + b = 1, then a4 + b4 – a3 – b3 – 2a2b2 + ab is equal to
1. 1
2. 2
3. 4
4. 0

1. a4 + b4 – a3 – b3 – 2a2b2 + ab
= a4 + b4 – 2a2b2 – a3 – b3 + ab
= (a + b)2 (a – b)2 – (a + b) (a2 – ab + b2) + ab
= (a – b)2 – a2 + ab – b2 + ab
[∵  a + b = 1]
= (a – b)2 – (a – b)2 = 0

##### Correct Option: D

a4 + b4 – a3 – b3 – 2a2b2 + ab
= a4 + b4 – 2a2b2 – a3 – b3 + ab
= (a + b)2 (a – b)2 – (a + b) (a2 – ab + b2) + ab
= (a – b)2 – a2 + ab – b2 + ab
[∵  a + b = 1]
= (a – b)2 – (a – b)2 = 0