If a+b+c=2, a²+b²+c²=6, a³+b³+c³=8, then a⁴+b⁴+c⁴ is?

Answer :

Given  :  a  +  b  +  c =  2                                                      ------------ (  A )
a² + b² + c² =  6                                                                    ------------ ( B )
And
a³ + b³ + c³ = 8                                                                     ------------ ( C )

Now we take whole square of equation (  A )  and get

( a  +  b +  c )²  = 2²

$\Rightarrow$a² + b² + c² + 2ab  + 2bc  + 2ca  =  4

$\Rightarrow$2 ( ab  + bc  + ca ) = 4 - ( a² + b² + c² ) , Substitute value from equation B , we get

$\Rightarrow$2 ( ab  + bc  + ca ) = 4 - 6 

$\Rightarrow$2 ( ab  + bc  + ca ) = - 2

$\Rightarrow$( ab  + bc  + ca ) = - 1                                                            ----------------- ( 1 )

Now

(  a  +  b + c ) ( a² + b² + c² )  = a³ + b³ + c³ + a²b + b²a +  b²c +c²b + c²a + a²c                                     --------------- (  2 )
And
(  a + b  + c ) ( ab  + bc + ca ) = a²b + b²a +  b²c +c²b + c²a + a²c   + 3abc  , So

$\Rightarrow$a²b + b²a +  b²c +c²b + c²a + a²c     = (  a + b  + c ) ( ab  + bc + ca ) - 3abc   , Substitute that value in equation 2 , we get

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

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

$\Rightarrow$a³ + b³ + c³ - 3abc = (  a  +  b + c )  [ ( a² + b² + c² )  - ( ab  + bc + ca ) ]

Now we substitute values from equation A , B , C and 1 , and get

$\Rightarrow$8  - 3abc  = 2 [ ( 6 ) - ( - 1 ) ]

$\Rightarrow$8 - 3abc  = 2 [ 7 ]

$\Rightarrow$- 3abc = 14 - 8

$\Rightarrow$-3abc  = 6

$\Rightarrow$abc  = - 2                                            ------------ (  3 )

Now we take whole square of equation 2 , and get

( a² + b² + c² ) ² = 6²

$\Rightarrow$( a⁴ + b⁴ + c⁴ ) + 2 ( a²b² + b²c² + c²a² )  = 36

$\Rightarrow$a⁴ + b⁴ + c⁴    = 36 - 2 ( a²b² + b²c² + c²a² )                      ----------- ( 4 )

And
( ab  + bc  + ca )² = a²b² + b²c² + c²a²  +  2 ( a²bc +ab²c + abc²  )  , So

$\Rightarrow$a²b² + b²c² + c²a²  = ( ab  + bc  + ca )²  - 2 ( a²bc +ab²c + abc²  ) 

$\Rightarrow$a²b² + b²c² + c²a²  = ( ab  + bc  + ca )²  - 2abc (  a + b  + c )  , Now we substitute values from equation A  ,  1 and 3 , we get 

$\Rightarrow$a²b² + b²c² + c²a²  = ( -1 )²  - 2 ( - 2 )(  2 ) 

$\Rightarrow$a²b² + b²c² + c²a²  = 1 + 8

$\Rightarrow$a²b² + b²c² + c²a²  = 9  , Substitute that value in equation 4 , we get

$\Rightarrow$a⁴ + b⁴ + c⁴    = 36 - 2 ( 9 )

$\Rightarrow$a⁴ + b⁴ + c⁴    = 36 - 18

$\Rightarrow$a⁴ + b⁴ + c⁴    = 18                                                                     ( Ans )