326. If x + y + z = 0, then the value of	x2 + y2 + z2	is
     	x2 - yz

     
     

1. 1. –1
      

   
   
   

   2. 0
      

   
   
   

   3. 1
      

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   x + y + z = 0
   ⇒ –x = y + z
   ⇒ (–x)² = (y + z)²
   ⇒ x² = y² + z² + 2yz ...(i)
   

   ∴ Expression =	x2 + y2 + z2
   	x2 - yz

   
   

   Expression =	y2 + z2 + 2yz + y2 + z2	{ ∴ Using (i) }
   	y2 + z2 + 2yz - yz

   
   

   Expression =	2y2 + 2z2 + 2yz
   	y2 + z2 + yz

   
   

   Expression =	2(y2 + z2 + yz)	= 2
   	y2 + z2 + yz

   
   

   Correct Option: D

   x + y + z = 0
   ⇒ –x = y + z
   ⇒ (–x)² = (y + z)²
   ⇒ x² = y² + z² + 2yz ...(i)
   

   ∴ Expression =	x2 + y2 + z2
   	x2 - yz

   
   

   Expression =	y2 + z2 + 2yz + y2 + z2	{ ∴ Using (i) }
   	y2 + z2 + 2yz - yz

   
   

   Expression =	2y2 + 2z2 + 2yz
   	y2 + z2 + yz

   
   

   Expression =	2(y2 + z2 + yz)	= 2
   	y2 + z2 + yz

   
   

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327. If a = 4.965, b = 2.343 and c = 2.622, then the value of a³ - b³ - c³ - 3abc is
     

1. 1. –2
      

   
   
   

   2. –1
      

   
   
   

   3. 0
      

   
   
   

   4. 9.93

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 21,
   a = 4.965, b = 2.343, c = 2.622
   a + (–b) + (–c) = 4.965 – 2.343 – 2.622 = 0
   ∴ a³ + b³ + c³ – 3 abc = a³ + (–b)³ + (–c)³ – 3abc = 0

   Correct Option: C

   Using Rule 21,
   a = 4.965, b = 2.343, c = 2.622
   a + (–b) + (–c) = 4.965 – 2.343 – 2.622 = 0
   ∴ a³ + b³ + c³ – 3 abc = a³ + (–b)³ + (–c)³ – 3abc = 0

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328. If a = 331, b = 336 and c = – 667, then the value of a³ + b³ + c³ - 3abc is
     

1. 1. 1
      

   
   
   

   2. 6
      

   
   
   

   3. 3
      

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 21,
   a + b + c = 331 + 336 – 667 = 0
   ∴ a³ + b³ + c³ – 3 abc = 0

   Correct Option: D

   Using Rule 21,
   a + b + c = 331 + 336 – 667 = 0
   ∴ a³ + b³ + c³ – 3 abc = 0

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329. If x +	1	= 2, then x2013 +	1		= ?
     	x		x2014

     
     

1. 1. 0
      

   
   
   

   2. 1
      

   
   
   

   3. –1
      

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   x +	1	= 2
   	x

   
   ⇒ x² + 1 = 2x
   ⇒ x² - 2x + 1 = 0
   ⇒ (x - 1)² = 0 ⇒ x = 1
   

   ∴ x2013 +	1	= 1 + 1 = 2
   	x2014

   
   Second Method :
   Using Rule 16,
   

   Here, x +	1	= 2
   	x

   
   

   ∴ x2013 +	1	= 2
   	x2014

   
   

   Correct Option: D

   x +	1	= 2
   	x

   
   ⇒ x² + 1 = 2x
   ⇒ x² - 2x + 1 = 0
   ⇒ (x - 1)² = 0 ⇒ x = 1
   

   ∴ x2013 +	1	= 1 + 1 = 2
   	x2014

   
   Second Method :
   Using Rule 16,
   

   Here, x +	1	= 2
   	x

   
   

   ∴ x2013 +	1	= 2
   	x2014

   
   

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330. If p =	5	, then 27p3 -	1	-	9	p2 +	1	p is equal to
     	18		216		2		4

     
     

1. 1. 	4
      	27

      
      

   
   
   

   2. 	5
      	27

      
      

   
   
   

   3. 	8
      	27

      
      

   
   
   

   4. 	10
      	27

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 9,
   

   We have ,	27p3 -	1	-	9	p2 +	1
   		216		2		4

   
   

   =	(3p)3 -		1		3	- 3.(3p)2		1		+ 3 × 3p ×	1	×	1
   			6					6			6		6

   
   

   =		3p -	1		3	=		3 ×	5	-	1		3
   			6						18		6

   
   

   =		5	-	1		3	=		4		3
   		6		6					6

   
   

   =		2		3	=	8
   		3				27

   
   

   Correct Option: C

   Using Rule 9,
   

   We have ,	27p3 -	1	-	9	p2 +	1
   		216		2		4

   
   

   =	(3p)3 -		1		3	- 3.(3p)2		1		+ 3 × 3p ×	1	×	1
   			6					6			6		6

   
   

   =		3p -	1		3	=		3 ×	5	-	1		3
   			6						18		6

   
   

   =		5	-	1		3	=		4		3
   		6		6					6

   
   

   =		2		3	=	8
   		3				27

   
   

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