Math, asked by angadbir15, 1 year ago

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

Answers

Answered by chaitu251798
7

a = 7  \\ b = 5 \\ c = 3 \\  7cube + 5cube + 3cube - 3 \times 7  \times 5 \times 3 \\ =  343 + 125 + 27 - 315 \\  =495  - 315 \\  = 180 \\ plz \: mark \: brainleist
Answered by Anonymous
34

Given:

  • a + b + c = 15,
  • a² + b² + c² = 83

To Find: Value of (a³ + b³ + c³ -3abc).

We know,

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

So, we have to fetch the value of (ab + bc + ca).

We also know,

(a + b + c)² = + + + 2(ab + bc + ca)

∴ (83) + 2(ab + bc + ca) = (15)²

→ 2(ab + bc + ca) = 225 - 83 = 142

→ ab + bc + ca = 142/2 = 71

NOW, putting the value:-

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

→ a³ + b³ + c³ - 3abc = (15)[83 - (ab + bc + ca)]

→ a³ + b³ + c³ - 3abc = (15)(83 - 71)

→ a³ + b³ + c³ - 3abc = (15)(12)

a³ + b³ + c³ - 3abc = 180

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