a+b+c=38 and a²+b²+c²=7202 and find the value of abc

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a+b+c=38 and a²+b²+c²=7202 and find the value of abc

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Eva 1 month 2021-08-14T02:55:05+00:00 1 Answer 0 views 0

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    2021-08-14T02:57:01+00:00

    Answer:

    I derived the following two relations:

    a³+b³+c³ -3abc = (a+b+c)³ -3(ab+bc+ca) (a+b+c)………………………………..……(1)

    (a+b+c)² = a² + b² + c² + 2 (ab+bc+ca)…………………………………………………….(2)

    Using the above two equations (1) and (2), we can find the answer to abc. We are given,

    a+b+c=6, a²+b²+c²=14 and a³+b³+c³=36……………………………..…………………(A)

    From eq.(2),

    2 (ab+bc+ca) = (a+b+c)² – (a² + b² + c²)

    = 6² – 14 = 36–14=22 [Using eq.(A)]

    Or, ab+bc+ca = 11………………………………………………………………………………(B)

    We now substitute the numerical values given in equations (A) and (B) into eq.(1) and obtain,

    36 – 3abc = 6³ – 3 x 11 x 6 = 216 – 198 = 18

    Or, -3abc = 18–36=-18 This yields

    abc = -18/-3 = 6 (Proved)

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