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

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

( 9)^2 = a ^2 + b ^2 + c ^2 + 2 ( ab + bc + ca)

SUBSTITUTING THE VALUES ,

81 = a^2 + b ^ 2 + c^2 + 2 (23)
81 - 46 = a^2 + b ^2 + c^2

SO ,

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