Question

If a+b+c=9 and a^2+b^2+c^2=35 find the value of (a^3+b^3+c^3-3abc)

Answer 1

Given (a + b + c) = 9
Squaring on both the sides, we get
(a + b + c)2 = 92
⇒ a2 + b2 + c2  + 2(ab + bc + ca) = 81
⇒ 35  + 2(ab + bc + ca) = 81
⇒ 2(ab + bc + ca) = 81 – 35 = 46
⇒ ab + bc + ca = 23  →  (1)
Recall that a3+b3+c3-3abc = (a + b + c)( a2 + b2 + c2 – ab – bc – ca)
                                       = 9(35 – 23)
                                       = 9(12) = 108

Answer 2

ANSWER

a+b+c=9 and a

2

+b

2

+c

2

=35

Using formula,

(a+b+c) 2

=a2+b 2+c 2+2(ab+bc+ca)

9 2=35+2(ab+bc+ca)

2(ab+bc+ca)=81−35=46

(ab+bc+ca)=23

using formula,

(a3+b3 +c3)−3abc=(a 2+b 2+c 2

-ab−bc−ca)(a+b+c)

a 3 +b 3+c 3

−3abc=(35−23)×9=9×12=108

Answer108