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    Question

    If a = 701, b= 703 and c= 706, then find the value of

    a³ + b³ +c³ –3abc?
    A 38230 Correct Answer Incorrect Answer
    B 40090 Correct Answer Incorrect Answer
    C 41120 Correct Answer Incorrect Answer
    D 42340 Correct Answer Incorrect Answer

    Solution

    Given – a= 701, b=703, c = 706 Then as we know that a³ + b³ +c³ –3abc = (a +b + c)/2 [(a-b) ² +(b-c) ² +(c-a) ²] = (701+703+706)/2 [4+9+25] =2110 / 2 [38] =2110×19= 40090

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