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Aptitude miscellaneous
  1. If a + b = 1 and a3 + b3 + 3ab = k, then the value of k is








  1. View Hint View Answer Discuss in Forum

    Using Rule 8,
    a + b = 1
    Cubing both sides,
    (a + b)3 = 13 = 1
    ⇒ a3 + b3 + 3ab(a + b ) = 1
    ⇒ a3 + b3 + 3ab = 1 = k
    ⇒ k = 1

    Correct Option: A

    Using Rule 8,
    a + b = 1
    Cubing both sides,
    (a + b)3 = 13 = 1
    ⇒ a3 + b3 + 3ab(a + b ) = 1
    ⇒ a3 + b3 + 3ab = 1 = k
    ⇒ k = 1

  1. If a = 34, b = c = 33, then the value of a3 + b3 + c3 – 3abc is








  1. View Hint View Answer Discuss in Forum

    Using Rule 22,

    a3 + b3 + c3 - 3abc =
    1
    		(a + b + c)[ (a - b)2 + (b - c)2 + (c - a)2 ]
    2

    Here , a = 34, b = c = 33
    ⇒ a3 + b3 + c3 - 3abc =
    1
    		(34 + 33 + 33)[ (34 - 33)2 + (33 - 33)2 + (33 - 34)2 ]
    2

    ⇒ a3 + b3 + c3 - 3abc =
    1
    		 × 100[ 1 + 0 + 1 ] = 100
    2

    Correct Option: D

    Using Rule 22,

    a3 + b3 + c3 - 3abc =
    1
    		(a + b + c)[ (a - b)2 + (b - c)2 + (c - a)2 ]
    2

    Here , a = 34, b = c = 33
    ⇒ a3 + b3 + c3 - 3abc =
    1
    		(34 + 33 + 33)[ (34 - 33)2 + (33 - 33)2 + (33 - 34)2 ]
    2

    ⇒ a3 + b3 + c3 - 3abc =
    1
    		 × 100[ 1 + 0 + 1 ] = 100
    2

  1. If a = √7 + 2√12 and b = √7 - 2√12 , then ( a3 + b3 ) is equal to








  1. View Hint View Answer Discuss in Forum

    a = √7 + 2 × √4 × √3
    = √4 + 3 + 2 × 2 × √3
    = √( 2 + √3 )2 = 2 + √3
    ∴ b = √7 - 2√12 = 2 - √3
    ⇒ a + b = 2 + √3 + 2 - √3 = 4
    ab = (2 + √3)(2 - √3) = 1
    ∴ a3 + b3 = ( a + b )3 - 3ab(a + b)
    = 64 – 3 × 4 = 52

    Correct Option: D

    a = √7 + 2 × √4 × √3
    = √4 + 3 + 2 × 2 × √3
    = √( 2 + √3 )2 = 2 + √3
    ∴ b = √7 - 2√12 = 2 - √3
    ⇒ a + b = 2 + √3 + 2 - √3 = 4
    ab = (2 + √3)(2 - √3) = 1
    ∴ a3 + b3 = ( a + b )3 - 3ab(a + b)
    = 64 – 3 × 4 = 52

  1. If a =
    b2
    		 then the value of a3 + b3 is
    b - a









  1. View Hint View Answer Discuss in Forum

    a =
    b2
    		⇒ ab - a2 = b2
    b - a

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

    Correct Option: B

    a =
    b2
    		⇒ ab - a2 = b2
    b - a

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

  1. If p = 99, then value of p(p2 + 3p + 3) is








  1. View Hint View Answer Discuss in Forum

    Using Rule 8
    Expression = p(p2 + 3p + 3)
    Expression = (p3 + 3p2 + 3p + 1) – 1
    Expression = (p + 1)3 – 1 = (99 + 1)3 – 1
    Expression = (100)3 – 1= 1000000 – 1 = 999999

    Correct Option: D

    Using Rule 8
    Expression = p(p2 + 3p + 3)
    Expression = (p3 + 3p2 + 3p + 1) – 1
    Expression = (p + 1)3 – 1 = (99 + 1)3 – 1
    Expression = (100)3 – 1= 1000000 – 1 = 999999