191. If p³ – q³ = (p – q) {(p – q)² + x p q)} then value of x is

1. 1. 1

   
   
   

   2. –1

   
   
   

   3. 3

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   a³ − b³ = (a – b) (a² + ab + b²)
   = (a – b) ((a + b)² – ab)
   ∴  On comparing with
   p³ – q³ = (p – q) ((p – q)² + x pq) x
   = 3

   Correct Option: C

   a³ − b³ = (a – b) (a² + ab + b²)
   = (a – b) ((a + b)² – ab)
   ∴  On comparing with
   p³ – q³ = (p – q) ((p – q)² + x pq) x
   = 3

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192. If a² = by + cz, b² = cz + ax, c² = ax + by, then the value of
     

     x	+	y	+	z	is
     a + x		b + y		c + z

1. 1. 1

   
   
   

   2. a + b + c

   
   
   

   3. 1	+	1	+	1
      a		b		c

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   a² = by + cz
   ⇒  a² + ax = ax + by + cz
   ⇒  a (a + x) = ax + by + cz
   

   ⇒	1	=	a
   	a + x		ax + by + cz

   
   Similarly,
   b² = cz + ax
   ⇒  b² + by = by + cz + ax
   ⇒  b (b + y) = ax + by + cz
   

   ⇒	1	=	b
   	b + y		ax + by + cz

   
   c² = ax + by
   ⇒  c² + cz = ax + by + cz
   ⇒  c (c + z) = ax + by + cz
   

   ⇒	1	=	c
   	c + z		ax + by + cz

   
   

   =	ax	+	by	+	cz
   	ax + by + cz		ax + by + cz		ax + by + cz

   
   

   =	ax + by + cz	= 1
   	ax + by + cz

   Correct Option: A

   a² = by + cz
   ⇒  a² + ax = ax + by + cz
   ⇒  a (a + x) = ax + by + cz
   

   ⇒	1	=	a
   	a + x		ax + by + cz

   
   Similarly,
   b² = cz + ax
   ⇒  b² + by = by + cz + ax
   ⇒  b (b + y) = ax + by + cz
   

   ⇒	1	=	b
   	b + y		ax + by + cz

   
   c² = ax + by
   ⇒  c² + cz = ax + by + cz
   ⇒  c (c + z) = ax + by + cz
   

   ⇒	1	=	c
   	c + z		ax + by + cz

   
   

   =	ax	+	by	+	cz
   	ax + by + cz		ax + by + cz		ax + by + cz

   
   

   =	ax + by + cz	= 1
   	ax + by + cz

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193. If p³ – q³ = (p – q) {(p + q)² – x p q} then the value of x is

1. 1. 1

   
   
   

   2. –1

   
   
   

   3. 2

   
   
   

   4. –2

   
   
   

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1. View Hint View Answer Discuss in Forum

   a³ − b³ = (a – b) (a² + ab + b²)
   = (a – b) {(a + b)² – ab}
   On comparing with
   p³ – q³ = (p – q) {(p + q)² – x pq)}, x
   = 1

   Correct Option: A

   a³ − b³ = (a – b) (a² + ab + b²)
   = (a – b) {(a + b)² – ab}
   On comparing with
   p³ – q³ = (p – q) {(p + q)² – x pq)}, x
   = 1

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194. If a – b = 1 and a³ – b³ = 61, then the value of ab will be

1. 1. –20

   
   
   

   2. 20

   
   
   

   3. 30

   
   
   

   4. 60

   
   
   

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1. View Hint View Answer Discuss in Forum

   a³ − b³
   = (a – b)³ + 3ab (a – b)
   ⇒  61 = 1 + 3ab × 1
   ⇒  61 – 1 = 3ab = 60
   

   ⇒  ab =	60	= 20
   	3

   Correct Option: B

   a³ − b³
   = (a – b)³ + 3ab (a – b)
   ⇒  61 = 1 + 3ab × 1
   ⇒  61 – 1 = 3ab = 60
   

   ⇒  ab =	60	= 20
   	3

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195. If	a	+	b	= 1, then the value of a3 + b3 will be
     	b		a

1. 1. 1

   
   
   

   2. 0

   
   
   

   3. –1

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   a	+	b	= 1
   b		a

   
   

   ⇒	a2 + b2	= 1
   	ab

   
   ⇒  a² + b² = ab
   ⇒  a² – ab + b² = 0
   ∴  a³ + b³ = (a + b) (a² – ab + b²)
   = 0

   Correct Option: B

   a	+	b	= 1
   b		a

   
   

   ⇒	a2 + b2	= 1
   	ab

   
   ⇒  a² + b² = ab
   ⇒  a² – ab + b² = 0
   ∴  a³ + b³ = (a + b) (a² – ab + b²)
   = 0

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