566. If x + y = a and xy = b² , then the value of x³ – x²y - xy² + y³ in terms of a and b is :
     

1. 1. (a² + 4b²)a
      

   
   
   

   2. a³ – 3b²
      

   
   
   

   3. a³ - 4b²a
      

   
   
   

   4. a³ + 3b²
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   x³ - x²y - xy² + y³ = x³ + y³ - x²y - xy²
   = (x + y)³ - 3xy(x + y) – xy(x + y)
   = (x + y)³ - 4xy(x + y) = a³ - 4b²a

   Correct Option: C

   x³ - x²y - xy² + y³ = x³ + y³ - x²y - xy²
   = (x + y)³ - 3xy(x + y) – xy(x + y)
   = (x + y)³ - 4xy(x + y) = a³ - 4b²a

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567. If x -	1	= 1 , then the value of		{ x4 - ( 1 / x2 ) }	is
     	x			3x2 + 5x - 3

1. 1. 	1
      	4

   
   
   

   2. 	1
      	2

   
   
   

   3. 	3
      	4

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Expression =	x4 -	1
   		x2
   	3x2 + 5x - 3

   
   Dividing numerator and denominator by x,
   
   
   

   Expression =		x3 -	1
   			x3
   		3x + 5 -	3
   			x

   
   
   

   Expression =		x3 -	1
   			x3
   	3		x -	1		+ 5
   				x

   
   

   Expression =		x -	1		3	+ 3		x -	1
   			x						x
   	3		x -	1		+ 5
   				x

   
   

   Expression =	1 + 3	=	4	=	1
   	3 + 5		8		2

   Correct Option: B

   Expression =	x4 -	1
   		x2
   	3x2 + 5x - 3

   
   Dividing numerator and denominator by x,
   
   
   

   Expression =		x3 -	1
   			x3
   		3x + 5 -	3
   			x

   
   
   

   Expression =		x3 -	1
   			x3
   	3		x -	1		+ 5
   				x

   
   

   Expression =		x -	1		3	+ 3		x -	1
   			x						x
   	3		x -	1		+ 5
   				x

   
   

   Expression =	1 + 3	=	4	=	1
   	3 + 5		8		2

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568. If x + y = 15, then (x – 10)³ + (y – 5)³ is
     

1. 1. 25
      

   
   
   

   2. 125
      

   
   
   

   3. 625
      

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   x + y = 15
   ⇒ (x – 10) + (y – 5) = 0
   ∴ (x – 10)³ + (y – 5)³ = (x – 10 + y – 5)³ – 3(x – 10)(y – 5)(x – 10 + y – 5) = 0
   [∴ a³ + b³ = (a + b)³ – 3ab (a + b)]
   

   Correct Option: D

   x + y = 15
   ⇒ (x – 10) + (y – 5) = 0
   ∴ (x – 10)³ + (y – 5)³ = (x – 10 + y – 5)³ – 3(x – 10)(y – 5)(x – 10 + y – 5) = 0
   [∴ a³ + b³ = (a + b)³ – 3ab (a + b)]
   

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569. If x2 +	1	= 66 , then the value of	x2 - 1 + 2x	= ?
     	x2		x

     
     

1. 1. ± 8
      

   
   
   

   2. 10, – 6
      

   
   
   

   3. 6, –10
      

   
   
   

   4. ± 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 5
   

   x2 +	1	= 66
   	x2

   
   

   ⇒		x -	1		2	+ 2 = 66
   			x

   
   

   ⇒		x -	1		2	= 66 - 2 = 64
   			x

   
   

   ⇒ x -	1	= ± 8
   	x

   
   

   ∴ Expression =	x2 - 1 + 2x
   	x

   
   

   Expression =	x2	-	1	+ 2 = x -	1	+ 2
   	x		x		x

   
   

   Putting the value of x -	1	= 8 + 2 or – 8 + 2 = 10 or –6
   	x

   Correct Option: B

   Using Rule 5
   

   x2 +	1	= 66
   	x2

   
   

   ⇒		x -	1		2	+ 2 = 66
   			x

   
   

   ⇒		x -	1		2	= 66 - 2 = 64
   			x

   
   

   ⇒ x -	1	= ± 8
   	x

   
   

   ∴ Expression =	x2 - 1 + 2x
   	x

   
   

   Expression =	x2	-	1	+ 2 = x -	1	+ 2
   	x		x		x

   
   

   Putting the value of x -	1	= 8 + 2 or – 8 + 2 = 10 or –6
   	x

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570. If a² + a + 1 = 0, then the value of a⁹ is
     

1. 1. 2
      

   
   
   

   2. 3
      

   
   
   

   3. 1
      

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 9,
   a² + a + 1 = 0
   ⇒ (a – 1) (a² + a + 1) = 0
   ⇒ a³ – 1 = 0
   ⇒ a³ = 1 ⇒ a = 1
   ∴ a⁹ = 1

   Correct Option: C

   Using Rule 9,
   a² + a + 1 = 0
   ⇒ (a – 1) (a² + a + 1) = 0
   ⇒ a³ – 1 = 0
   ⇒ a³ = 1 ⇒ a = 1
   ∴ a⁹ = 1

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