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

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

1. Answer:

   answer :

   hence , it is proved that LHS = RHS

   answer is 124

   hope it helps

   plz mark as brainliest

   have a great day

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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:

   [tex]\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}} [/tex]

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