1. If for two real constants a and b, the expression ax³ + 3x² - 8x + b is exactly divisible by (x + 2) and (x – 2), then

1. 1. a = 2, b = 12

   
   
   

   2. a = 12, b = 2

   
   
   

   3. a = 2, b = –12

   
   
   

   4. a = –2, b = 12

   
   
   

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1. View Hint View Answer Discuss in Forum

   P (x ) = ax³ + 3x² – 8x + b
   [∵ P (x) is div. by (x + 2) & (x –2)]
   ∴ P (–2) = –8a + 12 + 16 + b = 0
   ⇒ –8a + b + 28 = 0 ...(i)
   ⇒ P(2) = 8a + 12 – 16 + b = 2
   ⇒ 8a + b – 4 = 0 ...(ii)
   By equation (i) + (ii)
   2b + 24 = 0
   

   ⇒ b = -	24	= -12
   	2

   
   From equation (i),
   – 8a – 12 + 28 = 0
   ⇒ –8a = –16 ⇒ a = 2

   Correct Option: C

   P (x ) = ax³ + 3x² – 8x + b
   [∵ P (x) is div. by (x + 2) & (x –2)]
   ∴ P (–2) = –8a + 12 + 16 + b = 0
   ⇒ –8a + b + 28 = 0 ...(i)
   ⇒ P(2) = 8a + 12 – 16 + b = 2
   ⇒ 8a + b – 4 = 0 ...(ii)
   By equation (i) + (ii)
   2b + 24 = 0
   

   ⇒ b = -	24	= -12
   	2

   
   From equation (i),
   – 8a – 12 + 28 = 0
   ⇒ –8a = –16 ⇒ a = 2

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2. If x +	1	=	3	, find the value of 8x3 +	1
   	4x		2		8x3

1. 1. 18
      

   
   
   

   2. 36
      

   
   
   

   3. 24
      

   
   
   

   4. 16

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 8,
   

   x +	1	=	3
   	4x		2

   
   Multiplying both sides by 2
   

   ⇒ 2x +	1	= 3
   	2x

   
   On cubing both sides,
   

   ∴ 8x3 +	1	+	3 × 2x ×	1	×		2x +	1		= 27
   	8x3			2x				2x

   
   

   ⇒ 8x3 +	1	+ 3 × 3 = 27
   	8x3

   
   

   ⇒ 8x3 +	1	= 27 – 9 = 18
   	8x3

   Correct Option: A

   Using Rule 8,
   

   x +	1	=	3
   	4x		2

   
   Multiplying both sides by 2
   

   ⇒ 2x +	1	= 3
   	2x

   
   On cubing both sides,
   

   ∴ 8x3 +	1	+	3 × 2x ×	1	×		2x +	1		= 27
   	8x3			2x				2x

   
   

   ⇒ 8x3 +	1	+ 3 × 3 = 27
   	8x3

   
   

   ⇒ 8x3 +	1	= 27 – 9 = 18
   	8x3

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3. If x2 - 3x - 8x + 1 = 0, then the value of x3 +	1	is
   	x3

1. 1. 9
      

   
   
   

   2. 18
      

   
   
   

   3. 27
      

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 8,
   x² – 3x + 1 = 0
   ⇒ x² + 1 = 3x
   

   ⇒ x +	1	= 3
   	x

   
   

   ∴ x3 +

   1

   =

   x +

   1

   3

   - 3 × x ×

   1

   x +

   1

   x3

   x

   x

   x

   
   = 27 – 3 × 3 = 18
   

   Correct Option: B

   Using Rule 8,
   x² – 3x + 1 = 0
   ⇒ x² + 1 = 3x
   

   ⇒ x +	1	= 3
   	x

   
   

   ∴ x3 +

   1

   =

   x +

   1

   3

   - 3 × x ×

   1

   x +

   1

   x3

   x

   x

   x

   
   = 27 – 3 × 3 = 18
   

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4. If	1	=	1	+	1	( x ≠ 0,x ≠ 0 , x ≠ y ) then, the value of x3 - y3 is
   	x + y		x		y

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. -1

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   	1	=	1	+	1	=	y + x
   	x + y		x		y		xy

   
   ⇒ ( x + y )² = xy
   ⇒ x² + 2xy + y² = xy
   ⇒ x² + 2xy + y² - xy = 0
   ⇒ x² + xy + y² = 0
   ∴ x³ - y³ = (x – y)( x² + xy + y² ) = 0

   Correct Option: A

   	1	=	1	+	1	=	y + x
   	x + y		x		y		xy

   
   ⇒ ( x + y )² = xy
   ⇒ x² + 2xy + y² = xy
   ⇒ x² + 2xy + y² - xy = 0
   ⇒ x² + xy + y² = 0
   ∴ x³ - y³ = (x – y)( x² + xy + y² ) = 0

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5. If x +	1	= 2 ,find the value of 8x3 +		1	.
   	2x			x3

1. 1. 48

   
   
   

   2. 88

   
   
   

   3. 40

   
   
   

   4. 44

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 8,
   

   x +	1	= 2
   	2x

   
   

   ⇒ 2x +	2	= 2 × 2 = 4
   	2x

   
   

   ⇒ 2x +	1	= 4
   	x

   
   On cubing both sides,
   

   ∴ 8x3 +	1	+	3 × 2x ×	1		2x +	1		= 64
   	x3			x			x

   
   

   ⇒ 8x3 +	1	+ 6 × 4 = 64
   	x3

   
   

   ⇒ 8x3 +	1	= 64 – 24 = 40
   	x3

   Correct Option: C

   Using Rule 8,
   

   x +	1	= 2
   	2x

   
   

   ⇒ 2x +	2	= 2 × 2 = 4
   	2x

   
   

   ⇒ 2x +	1	= 4
   	x

   
   On cubing both sides,
   

   ∴ 8x3 +	1	+	3 × 2x ×	1		2x +	1		= 64
   	x3			x			x

   
   

   ⇒ 8x3 +	1	+ 6 × 4 = 64
   	x3

   
   

   ⇒ 8x3 +	1	= 64 – 24 = 40
   	x3

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