Question

If a + b + c = 5 and ab + bc + ca = 10, then prove that a
3 + b
3 + c
3 –3abc = – 25

Answer 1

Given----> a + b + c = 5 and ab + bc + ca = 10

To prove ----> a³ + b³ + c³ - 3abc = - 25

Proof----> We know that,

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

= ( a² + b² + c² ) + 2 ( ab + bc + ca )

Putting ( a + b + c ) = 5 and ab + bc + ca = 10 , we get,

=> ( 5 )² = ( a² + b² + c² ) + 2 ( 10 )

=> 25 = ( a² + b² + c² ) + 20

=> ( a² + b² + c² ) = 25 - 20

=> ( a² + b² + c² ) = 5

Now, we know that,

( a³ + b³ + c³ - 3abc )

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

Putting ( a + b + c ) = 5 , we get,

= ( 5 ) { ( a² + b² + c² ) - ( ab + bc + ca ) }

Putting , ( a² + b² + c² ) = 5 and ( ab + bc + ca ) = 10 , we get,

= ( 5 ) { ( 5 ) - ( 10 ) }

= ( 5 ) ( 5 - 10 )

= ( 5 ) ( - 5 )

= - 25