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

Hi ,

a + b + c= 9 ------( 1 )

ab + bc + ca = 23-------( 2 )

multiply both sides of equation( 2 )with 2 ,

2ab + 2bc + 2ca = 46 -------( 3 )

As we know the algebraic identity ,


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

a² + b² + c² + 46 = 9²    { from(1 ) and (3 ) }

a² + b² + c² = 81 - 46

                   = 35

I hope this helps you .

: )