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