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Aptitude miscellaneous
  1. If a + b + c = 15 and a2 + b2 + c2 = 83 then the value of a3 + b3 + c3 – 3abc








  1. View Hint View Answer Discuss in Forum

    a + b + c = 15
    ∴ (a + b + c)2 = 225
    ∴ a2 + b2 + c2 + 2 (ab + bc + ca)
    = 225
    ⇒ 2 (ab + bc + ca) = 225 – 83
    = 142
    ⇒ ab + bc + ca = 142 ÷ 2 = 71
    ∴ a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
    = 15 (83 – 71) = 15 × 12 = 180

    Correct Option: B

    a + b + c = 15
    ∴ (a + b + c)2 = 225
    ∴ a2 + b2 + c2 + 2 (ab + bc + ca)
    = 225
    ⇒ 2 (ab + bc + ca) = 225 – 83
    = 142
    ⇒ ab + bc + ca = 142 ÷ 2 = 71
    ∴ a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
    = 15 (83 – 71) = 15 × 12 = 180

  1. . If a – b = 3 and a3 - b3 = 117 then |a + b| is equal to








  1. View Hint View Answer Discuss in Forum

    a – b = 3
    a3 - b3 = 117
    a3 - b3 = ( a - b )3 + 3ab (a – b)
    ⇒ 117 = 27 + 3ab (3)
    ⇒ 9ab = 117 – 27 = 90
    ⇒ ab = 10
    ∴ ( a + b )2 = ( a - b )2 + 4ab
    = 9 + 40 = 49
    ∴ |a + b| = 7

    Correct Option: C

    a – b = 3
    a3 - b3 = 117
    a3 - b3 = ( a - b )3 + 3ab (a – b)
    ⇒ 117 = 27 + 3ab (3)
    ⇒ 9ab = 117 – 27 = 90
    ⇒ ab = 10
    ∴ ( a + b )2 = ( a - b )2 + 4ab
    = 9 + 40 = 49
    ∴ |a + b| = 7

  1. If x +
    1
    		 = 1 ,then ( x + 1 )5 +
    1
    		 equals
    x + 1( x + 1 )5









  1. View Hint View Answer Discuss in Forum

    x +
    1
    		 = 1
    x + 1

    Adding 1 on both sides ,
    ⇒ ( x + 1 ) +
    1
    		 = 2
    ( x + 1 )

    On squaring,
    ⇒ ( x + 1 )2 +
    1
    		 + 2 = 4
    ( x + 1 )2

    ⇒ ( x + 1 )2 +
    1
    		 = 2 ...(i)
    ( x + 1 )2

    Again, cubing , ( x + 1 ) +
    1
    		 = 2
    ( x + 1 )

    ⇒ ( x + 1 )3 +
    1
    		 + 3( x + 1 ) +
    1
    		 = 8
    ( x + 1 )3( x + 1 )

    ⇒ ( x + 1 )3 +
    1
    		 = 8 – 3 × 2 = 2
    ( x + 1 )3

    ∴ ( x + 1 )2 +
    1
    		( x + 1 )3 +
    1
    		 = 2 × 2 = 4
    ( x + 1 )2( x + 1 )3

    ⇒ ( x + 1 )5 + ( x + 1 ) +
    1
    		 +
    1
    		 = 4
    ( x + 1 )( x + 1 )5

    ⇒ ( x + 1 )5 +
    1
    		 = 4 – 2 = 2
    ( x + 1 )5

    Second method :
    Here, x +
    1
    		 = 1
    x + 1

    ⇒ ( x + 1 ) +
    1
    		 = 2
    ( x + 1 )

    ∴ ( x + 1 )2 +
    1
    		 = 2
    ( x + 1 )2

    Correct Option: B

    x +
    1
    		 = 1
    x + 1

    Adding 1 on both sides ,
    ⇒ ( x + 1 ) +
    1
    		 = 2
    ( x + 1 )

    On squaring,
    ⇒ ( x + 1 )2 +
    1
    		 + 2 = 4
    ( x + 1 )2

    ⇒ ( x + 1 )2 +
    1
    		 = 2 ...(i)
    ( x + 1 )2

    Again, cubing , ( x + 1 ) +
    1
    		 = 2
    ( x + 1 )

    ⇒ ( x + 1 )3 +
    1
    		 + 3( x + 1 ) +
    1
    		 = 8
    ( x + 1 )3( x + 1 )

    ⇒ ( x + 1 )3 +
    1
    		 = 8 – 3 × 2 = 2
    ( x + 1 )3

    ∴ ( x + 1 )2 +
    1
    		( x + 1 )3 +
    1
    		 = 2 × 2 = 4
    ( x + 1 )2( x + 1 )3

    ⇒ ( x + 1 )5 + ( x + 1 ) +
    1
    		 +
    1
    		 = 4
    ( x + 1 )( x + 1 )5

    ⇒ ( x + 1 )5 +
    1
    		 = 4 – 2 = 2
    ( x + 1 )5

    Second method :
    Here, x +
    1
    		 = 1
    x + 1

    ⇒ ( x + 1 ) +
    1
    		 = 2
    ( x + 1 )

    ∴ ( x + 1 )2 +
    1
    		 = 2
    ( x + 1 )2

  1. If
    1
    		 -
    1
    		 =
    1
    		 , then the value of a3 + b3 is
    aba - b









  1. View Hint View Answer Discuss in Forum

    ∴
    1
    		 +
    1
    		 =
    1
    aba - b

    ⇒
    b - a
    		 =
    1
    aba - b

    ⇒ (a – b) (a – b) = –ab
    ⇒ a2 – 2ab + b2 = –ab
    ⇒ a2 – ab + b2 = 0
    ∴ a3 + b3 = (a + b)(a2 – ab + b2) = 0

    Correct Option: A

    ∴
    1
    		 +
    1
    		 =
    1
    aba - b

    ⇒
    b - a
    		 =
    1
    aba - b

    ⇒ (a – b) (a – b) = –ab
    ⇒ a2 – 2ab + b2 = –ab
    ⇒ a2 – ab + b2 = 0
    ∴ a3 + b3 = (a + b)(a2 – ab + b2) = 0

  1. If a + b + c = 0, then a3 + b3 + c3 is equal to








  1. View Hint View Answer Discuss in Forum

    Using Rule 21,
    ∴ a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
    If a + b + c = 0, then a3 + b3 + c3 = 3abc

    Correct Option: D

    Using Rule 21,
    ∴ a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
    If a + b + c = 0, then a3 + b3 + c3 = 3abc