Algebra


  1. If
    p2
    +
    q2
    = 1 , then the value of (p6 + q6) is
    q2p2










  1. View Hint View Answer Discuss in Forum

    p2
    +
    q2
    = 1
    q2p2

    p4 + q4
    = 1
    p2q2

    ⇒ p4 + q4 = p2q2
    ⇒ p4 + q4 - p2q2 = 0 ...... (i)
    ∴ p6 + q6 = (p2)3 + (q2)3
    p6 + q6 = (p2 + q2)(p4 + q4 - p2q2)
    [ ∴ a3 + b3 = (a + b)(a2 + b2 - ab) ]
    p6 + q6 = (p2 + q2) × 0 = 0

    Correct Option: A

    p2
    +
    q2
    = 1
    q2p2

    p4 + q4
    = 1
    p2q2

    ⇒ p4 + q4 = p2q2
    ⇒ p4 + q4 - p2q2 = 0 ...... (i)
    ∴ p6 + q6 = (p2)3 + (q2)3
    p6 + q6 = (p2 + q2)(p4 + q4 - p2q2)
    [ ∴ a3 + b3 = (a + b)(a2 + b2 - ab) ]
    p6 + q6 = (p2 + q2) × 0 = 0


  1. If a + b – c = 0 then the value of 2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – c4









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    Expression = 2b2c2 + 2c2a2 + 2a2b2 - a4 - b4 - c4
    Expression = 4b2c2 - (2b2c2 - 2c2a2 - 2a2b2 + a4 + b4 + c4)
    Expression = (2bc)2 - (a2 - b2 - c2)2
    Expression = (2bc + a2 - b2 - c2)(2bc - a2 + b2 + c2)
    Expression = {a2 - (b2 + c2 - 2bc){(b2 + c2 + 2bc) - a2}
    Expression = { a2 - (b - c)2 }{ (b + c)2 - a2 }
    Expression = (a – b + c) (a + b – c)(a + b + c) (b + c – a)
    If a + b – c = 0,
    ∴ Expression = 0

    Correct Option: B

    Expression = 2b2c2 + 2c2a2 + 2a2b2 - a4 - b4 - c4
    Expression = 4b2c2 - (2b2c2 - 2c2a2 - 2a2b2 + a4 + b4 + c4)
    Expression = (2bc)2 - (a2 - b2 - c2)2
    Expression = (2bc + a2 - b2 - c2)(2bc - a2 + b2 + c2)
    Expression = {a2 - (b2 + c2 - 2bc){(b2 + c2 + 2bc) - a2}
    Expression = { a2 - (b - c)2 }{ (b + c)2 - a2 }
    Expression = (a – b + c) (a + b – c)(a + b + c) (b + c – a)
    If a + b – c = 0,
    ∴ Expression = 0



  1. If 4a -
    4
    + 3 = 0 then the value of : a3 -
    1
    + 3 = ?
    aa3










  1. View Hint View Answer Discuss in Forum

    4a -
    4
    = -3
    a

    On dividing by 4,
    ⇒ a -
    1
    =
    -3
    a4

    ∴ a3 -
    1
    = a -
    1
    3 + 3 . a .
    1
    a -
    1
    a3aaa

    ⇒ a3 -
    1
    =
    -3
    3 + 3 ×
    -3
    a344

    ⇒ a3 -
    1
    = -
    27
    -
    9
    a3644

    ⇒ a3 -
    1
    =
    -27 - 144
    =
    -171
    a36464

    ∴ a3 -
    1
    + 3 =
    -171
    + 3 =
    -171 + 192
    a36464

    a3 -
    1
    + 3 =
    21
    a364

    Correct Option: C

    4a -
    4
    = -3
    a

    On dividing by 4,
    ⇒ a -
    1
    =
    -3
    a4

    ∴ a3 -
    1
    = a -
    1
    3 + 3 . a .
    1
    a -
    1
    a3aaa

    ⇒ a3 -
    1
    =
    -3
    3 + 3 ×
    -3
    a344

    ⇒ a3 -
    1
    = -
    27
    -
    9
    a3644

    ⇒ a3 -
    1
    =
    -27 - 144
    =
    -171
    a36464

    ∴ a3 -
    1
    + 3 =
    -171
    + 3 =
    -171 + 192
    a36464

    a3 -
    1
    + 3 =
    21
    a364


  1. If x2 + x = 5 then the value of (x + 3)3 +
    1
    is
    (x + 3)3










  1. View Hint View Answer Discuss in Forum

    x2 + x = 5 (Given)
    Let, x + 3 = a

    1
    =
    1
    x + 3a

    Now , a +
    1
    = (x + 3) +
    1
    a(x + 3)

    a +
    1
    =
    (x + 3)2 + 1
    a(x + 3)

    a +
    1
    =
    x2 + 6x + 9 + 1
    ax + 3

    a +
    1
    =
    x2 + x + 5x + 10
    ax + 3

    a +
    1
    =
    5 + 5x + 10
    ax + 3

    a +
    1
    =
    5x + 15
    =
    5(x + 3)
    = 5
    ax + 3x + 3

    ∴ a3 +
    1
    = a +
    1
    3 - 3 . a .
    1
    a +
    1
    a3aaa

    Required answer = (5)3 – 3 × 5 = 125 – 15 = 110

    Correct Option: B

    x2 + x = 5 (Given)
    Let, x + 3 = a

    1
    =
    1
    x + 3a

    Now , a +
    1
    = (x + 3) +
    1
    a(x + 3)

    a +
    1
    =
    (x + 3)2 + 1
    a(x + 3)

    a +
    1
    =
    x2 + 6x + 9 + 1
    ax + 3

    a +
    1
    =
    x2 + x + 5x + 10
    ax + 3

    a +
    1
    =
    5 + 5x + 10
    ax + 3

    a +
    1
    =
    5x + 15
    =
    5(x + 3)
    = 5
    ax + 3x + 3

    ∴ a3 +
    1
    = a +
    1
    3 - 3 . a .
    1
    a +
    1
    a3aaa

    Required answer = (5)3 – 3 × 5 = 125 – 15 = 110



  1. If x2 + y2 + z2 = 2 (x + z – 1), then the value of :
    x3 + y3 + z3 = ?









  1. View Hint View Answer Discuss in Forum

    x2 + y2 + z2 = 2(x + z – 1)
    ⇒ x2 + y2 + z2 = 2x + 2z – 2
    ⇒ x2 - 2x + y2 + z2 - 2z + 2 = 0
    ⇒ x2 - 2x + 1 + y2 + z2 - 2z + 1 = 0
    ⇒ (x - 1)2 + y2 + (z - 1)2 = 0
    ∴ a2 + b2 + c2 = 0 ⇒ a = 0, b = 0, c = 0]
    ∴ x – 1 = 0 ⇒ x = 1
    y = 0
    z – 1 = 0 ⇒ z = 1
    ∴ x3 + y3 + z3 = 1 + 0 + 1 = 2

    Correct Option: A

    x2 + y2 + z2 = 2(x + z – 1)
    ⇒ x2 + y2 + z2 = 2x + 2z – 2
    ⇒ x2 - 2x + y2 + z2 - 2z + 2 = 0
    ⇒ x2 - 2x + 1 + y2 + z2 - 2z + 1 = 0
    ⇒ (x - 1)2 + y2 + (z - 1)2 = 0
    ∴ a2 + b2 + c2 = 0 ⇒ a = 0, b = 0, c = 0]
    ∴ x – 1 = 0 ⇒ x = 1
    y = 0
    z – 1 = 0 ⇒ z = 1
    ∴ x3 + y3 + z3 = 1 + 0 + 1 = 2