Question

A+b+c=9and ab+bc+ca=23 find a²+b²+c²

Answer 1

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

a+b+c= 9 (given)

ab+ bc + ca = 23 (given)

⇒ a³+b³+c³- 3abc = 9 (a²+b²+c²- 23)            ... (1)

 (a+b+c)²= a²+b²+c²+ 2ab+ 2bc+ 2ca

92= a²+b²+c²+ 2(ab+bc+ca)

81= a²+b²+c²+ 2(23)

81 = a²+b²+c²+ 46

a²+b²+c²= 35

 from eqation 1 we get,

a³+b³+c³- 3abc 

= 9 (35-23) 

= 108

Answer 2

Given that a+b+c = 9 and ab+bc+ca=23.

Given a+b+c=9

On squaring both sides, we get

(a+b+c)^2 = (9)^2

a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81

a^2 + b^2 + c^2 + 2(23) = 81

a^2 + b^2 + c^2 + 46 = 81

a^2 + b^2 + c^2 = 81 - 46

a^2 + b^2 + c^2 = 35.


Hope this helps!