Question

a+b+c =9 and ab +bc+ca=40 find a squate +b square +c square

Answer 1

Hi ,

It is given that ,

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

ab + bc + ca = 40

2( ab + bc + ca ) = 80

2ab+ 2bc + 2ca = 80 ----( 2 )

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We know the algebraic identity ,

(a+b+c)² = a² + b² + c² + 2ab+2bc+2ca

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a²+ b²+c² = ( a+b+c)² - (2ab+2bc+2ca)

= 9² - 80

[ From ( 1 ) & ( 2 ) ]

= 81 - 80

= 1

I hope this helps you.

: )

Answer 2

a+b+c =9

Square on both sides,

(a+b+c)²=9²

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

Here, a²+b²+c²+2(a+bc+ca) =9

Put the given value

a²+b²+c²+(2*40)=81

a²+b²+c²+80=81

a²+b²+c²=81-81

a²+b²+c²=1



I hope this will help you



-by ABHAY