MathematicsIf a+b+c=5 and ab+bc+ca=10, then the value of a3+b3+c3-3abc is

If a+b+c=5 and ab+bc+ca=10, then the value of a3+b3+c3-3abc is


  1. A
    25
  2. B
    -25
  3. C
    35
  4. D
    -35 

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    Solution:

    It is given that a+b+c=5 and ab+bc+ca=10.
    We know a+b+c2=a2+b2+c2+2ab+bc+ca.
    Substituting values, a+b+c=5 and ab+bc+ca=10 in the above identity,
    52=a2+b2+c2+2(10)
    25=a2+b2+c2+20
    25-20=a2+b2+c2
    5=a2+b2+c2
    a2+b2+c2=5          …….eq(i)
    Now, we know a3+b3+c3-3abc=(a+b+c)a2+b2+c2-ab-bc-ca. Substituting values, a+b+c=5, ab+bc+ca=10 and a2+b2+c2=5 in above identity,
    a3+b3+c3-3abc=(a+b+c)a2+b2+c2-ab-bc-ca
    a3+b3+c3-3abc=(5)(5-10)
    a3+b3+c3-3abc=(5)(-5)
    a3+b3+c3-3abc=-25
    Therefore, option 2 is correct.
     
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