Question

a + b + c is equals to 15 and a square B square + c square is equal to 83 find the value of a cube plus b cube plus c cube minus 3 ABC

Answer 1

(a+b+c) ^2 =a^2 +b^2 +c^2 +2(ab+bc+ca) 15^2 =83+2(ab+bc+ca) (225-83)/2=ab+bc+ca 142/2=71=ab+bc+ca (a+b+c) ^3 =a^3+ b^3+c^3+(a+b+c) (a2+b2+c2+2(ab+bc+ca) a3+b3+c3=(a+b+c)((a+b+c)^2 - 2(a2+b2+c2+2(ab+bc+ca)) =(15)((15)^2 - 2(83+2(71))) 15*(225-2*225) =-15*225=  - 3375

Answer 2

Answer:

180

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