Question

If a+b+c=9
and ab+ be+ca=23,
then a3 + b3 +c3-3abc =​

Answer 1

hope it was helpful to you ☺️☺️

mark me brainlist ❣️❣️

Answer 2

Answer:

108

Step-by-step explanation:

Given,

a + b + c = 9

ab + bc + ca = 23

We know that, ( a + b + c )³ = ( a + b + c ) ( a² + b² + c² - ( ab + bc + ca ) )

Using the algebraic identity, ( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

⇒ a² + b² + c² + 2( ab + bc + ca ) = ( 9 )²

   a² + b² + c² + 2( 23 ) = 81

   a² + b² + c² + 46 = 81

   a² + b² + c² = 81 -46

  a² + b² + c² = 35

Now finding out the value of a³ + b³ + c³ - 3abc

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

  a³ + b³ + c³ - 3abc = ( 9 ) [ ( 35 ) - ( 23 ) ]

  a³ + b³ + c³ - 3abc = ( 9 ) ( 12 )

  a³ + b³ + c³ - 3abc = 108

∴ The value of a³ + b³ + c³ - 3abc is 108.

[ Main points are written in bold form and underlined for your better understanding ]