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ICSE-VIII-Mathematics

12: Algebraic Identities Class 8 Maths

with Solutions -
Qstn# A-3-i Prvs-QstnNext-Qstn
  • #3-i
    (a + 1)(a - 1)(a2 + 1) (ii) (a + b)(a - b)(a2 + b2) (iii) (2a - b)(2a + b)(4a2 + b2) (iv) (3 - 2x)(3 + 2x)(9 + 4x2) (v) (3x - 4y)(3x + 4y)(9x2 + 16y2)
    Ans : (a + 1)(a - 1)(a2 + 1)
    = [
    (a)2 - (1)2] (a2 + 1)
    = (a2 - 1)(a2 + 1)
    = (a2)2 - (1)2
    = a4 - 1 (ii) (a + b)(a - b)(a2 + b2)
    = (a2 - b2)(a2 + b2)
    = (a2 - 1)(a2 + 1)
    = (a2)2 - (1)2
    = a4 - 1 (iii) (2a - b)(2a + b)(4a2 + b2)
    = [(2a)2 -
    (b)2] (4a2 + b2)
    = (4a2 - b2)(4a2 + b2)
    = (4a2)2 - (b2)2
    = (16a4 - b4) (iv) (3 - 2x)(3 + 2x)(9 + 4x2)
    = [(3)2 - (2x)2] (9 + 4x2)
    = (9 - 4x2)(9 + 4x2)
    = (9)2 - (4x2)2
    = 81 - 16x4 (v) (3x - 4y)(3x + 4y)(9x2 + 16y2)
    = [(3x)2 - (4y)2](9x2 + 16y2)
    = (9x2 - 16y2)(9x2 + 16y2)
    = (9x2)2 - (16y2)2
    = 81x4 - 256y4
  • #3-ii
    (a + b)(a - b)(a2 + b2)
    Ans : (a + b)(a - b)(a2 + b2)
    = (a2 - b2)(a2 + b2)
    = (a2 - 1)(a2 + 1)
    = (a2)2 - (1)2
    = a4 - 1
  • #3-iii
    (2a - b)(2a + b)(4a2 + b2)
    Ans : (2a - b)(2a + b)(4a2 + b2)
    = [(2a)2 -
    (b)2] (4a2 + b2)
    = (4a2 - b2)(4a2 + b2)
    = (4a2)2 - (b2)2
    = (16a4 - b4)
  • #3-iv
    (3 - 2x)(3 + 2x)(9 + 4x2)
    Ans : (3 - 2x)(3 + 2x)(9 + 4x2)
    = [(3)2 - (2x)2] (9 + 4x2)
    = (9 - 4x2)(9 + 4x2)
    = (9)2 - (4x2)2
    = 81 - 16x4
  • #3-v
    (3x - 4y)(3x + 4y)(9x2 + 16y2)
    Ans : (3x - 4y)(3x + 4y)(9x2 + 16y2)
    = [(3x)2 - (4y)2](9x2 + 16y2)
    = (9x2 - 16y2)(9x2 + 16y2)
    = (9x2)2 - (16y2)2
    = 81x4 - 256y4