Question

If a+ b +c= 1 and a2 +b2 +c2= 83 find value of a3 + b3 +c3 -3abc

Answer 1

Answer:

answer :

hence , it is proved that LHS = RHS

answer is 124

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Answer 2

ANSWER

Given:

• a + b + c = 1
• a² + b² + c² = 83

To Find:

• Value of a³ + b³ + c³ - 3abc

Solution:

We are given that,

⇒ a² + b² + c² = 83------------(1)

⇒ a + b + c = 1 --------(2)

So,

⇒ c = 1 - a - b --------(3)

Putting the value of 'c' from (3) into (1)

⇒ a² + b² + (1 - a - b)² = 83

⇒ a² + b² + (1 + a² + b² - 2a - 2b + 2ab ) = 83

⇒ a² + b² + a² + b² - 2a - 2b + 2ab = 82

⇒ 2a² + 2b² - 2a - 2b + 2ab = 82

⇒ a² + b² - a - b + ab = 41-----------(4)

We know that,

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

From (1), (2) and (3)

⇒ a³ + b³ + c³ - 3abc = (1)[83+ (- ab - b(1 - a - b) - (1 - a - b)a)]

⇒ a³ + b³ + c³ - 3abc = 83 + (- ab - b + ab + b² - a + a² + ab)

⇒ a³ + b³ + c³ - 3abc = 83 + (a²+ b² - a - b + ab)

From (4),

⇒ a³ + b³ + c³ - 3abc = 83 + 41

⇒ a³ + b³ + c³ - 3abc = 124

Formula Used:

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

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$