Math, asked by Firerage6347, 3 days ago

c²-2ab-a²-b²,a²+b²+c³-3abc,b²-2ca-c²-a²

Answers

Answered by jitendragurav097
0

Step-by-step explanation:

Given, a + b + c = 15 and a² + b² + c² = 83

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

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

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

⇒ 2(ab + bc + ca) = (15)² - 83 = 225 - 83 = 142

⇒ ab + bc + ca = 71

⇒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] = 180

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