Aptitude question of SAP Labs, answer:
Ques. Prove that (1/a3) + (1/b3) + (1/c3) = (1/a + 1/b + 1/c)3 + 3/abc.
  • Kundan Sharma
  • Kundan
  • 22 sep.
  • 515 Visto
  • 3 Respuestas

3 Respuestas
1-3 of  3
3 Respuestas
  • We know the identity a3 + b3 + c3 = (a + b + c)3 – 3(a + b)(b + c)(c + a),. For a + b + c = 0 ⇒ a + b = – c, b + c = – a, c + a = – b. Then (1/a)3 + (1/b)3 + (1/c)3 = (1/a + 1/b + 1/c)3 – 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 – 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 – 3(abc / a2b2c2) = (1/a + 1/b + 1/c)3 – 3 / abc.(proved)


  • a3 + b3 + c3 = (a + b + c)3 – 3(a + b)(b + c)(c + a)

    a + b + c = 0 ⇒ a + b = – c, b + c = – a, c + a = – b.

    (1/a)3 + (1/b)3 + (1/c)3
    = (1/a + 1/b + 1/c)3 – 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a)
    = (1/a + 1/b + 1/c)3 – 3(a+b / ab)(b + c / bc)(c + a / ac)
    = (1/a + 1/b + 1/c)3 – 3(– c / ab)(– a / bc)(– b / ac)
    = (1/a + 1/b + 1/c)3 – 3(abc / a2b2c2)
    = (1/a + 1/b + 1/c)3 – 3 / abc.
    Hence Proved


  • dnt knw

    Vote up -1 Vote down

SAP LABS
Practica simulacro de prueba para
SAP LABS