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

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

  • 94

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

(15)2=83+2(ab+bc+ca) 

225= 83+2(ab+bc+ca)

142=2(ab+bc+ca)

ab+bc+ca=71

  • 13

 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

  • 25
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