If a + b + c = 15 and a²+ b²+ c²= 83, find the value of a³+ b³+ c³–3abc.

Question

If a + b + c = 15 and a²+ b²+ c²= 83, find the value of a³+ b³+ c³–3abc.

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Josie 1 month 2021-08-18T10:04:05+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-18T10:05:29+00:00

    Given :

    • a + b + c = 15

    • a² + b² + c² = 83

    To Find :

    • a³ + b³ + c³ – 3abc

    Solution :

    We know that,

    ★ a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

    → a³ + b³ + c³ – 3abc = (a + b + c) {(a² + b² + c²) – ab + bc + ca)} … i)

    From the above identity, we’re required to find the values of a + b + c, + + and ab + bc + ca to get the value of + + 3abc

    Now,

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

    → 15² = 83 + 2(ab + bc + ca)

    → 225 = 83 + 2(ab + bc + ca)

    → 225 – 83 = 2(ab + bc + ca)

    → 142 = 2(ab + bc + ca)

    ab + bc + ca = 71

    Put the value of ab + bc + ca in eq i)

    ★ a³ + b³ + c³ – 3abc = (a + b + c) {(a² + b² + c²) – ab + bc + ca)}

    → a³ + b³ + c³ – 3abc = 15 × (83 – 71)

    → a³ + b³ + c³ – 3abc = 15 × 12

    a³ + b³ + c³ – 3abc = 180

    \therefore The value of a³ + b³ + c³ – 3abc is 180.

    0
    2021-08-18T10:05:37+00:00

    Answer:

    hey mate here is ur answer

    Step-by-step explanation:

    a³ + b³ + c³ – 3abc = ( a + b + c ) ( a² + b² + c² – ( ab + bc + ca ). = ( 15 ) ( 83 – ( 71 ) ). = 15 × 12. = 180

    please mark as brainleast

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