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

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

$⇒$a2 + b2 + c2 + 2ab  + 2bc  + 2ca  =  4

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

$⇒$2 ( ab  + bc  + ca ) = 4 - 6

$⇒$2 ( ab  + bc  + ca ) = - 2

$⇒$( 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

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

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

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

$⇒$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

$⇒$8  - 3abc  = 2 [ ( 6 ) - ( - 1 ) ]

$⇒$8 - 3abc  = 2 [ 7 ]

$⇒$- 3abc = 14 - 8

$⇒$-3abc  = 6

$⇒$abc  = - 2                                            ------------ (  3 )

Now we take whole square of equation 2 , and get

( a2 + b2 + c2 ) 2 = 62

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

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

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

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

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

$⇒$a2b2 + b2c2 + c2a2  = ( -1 )2  - 2 ( - 2 )(  2 )

$⇒$a2b2 + b2c2 + c2a2  = 1 + 8

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

$⇒$a4 + b4 + c4    = 36 - 2 ( 9 )

$⇒$a4 + b4 + c4    = 36 - 18

$⇒$a4 + b4 + c4    = 18                                                                     ( Ans )

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