Question

if a+b+c=1, a²+b²+c²=2 and a³+b³+c³=3 then prove that a⁴+b⁴+c⁴=4⅙​

Answer 1

Given : a+b+c = 1

a²+b²+c² = 2

a³+b³+c³ = 3

To Find : prove that  a⁴+b⁴+c⁴ =  4⅙​  

Solution:

a²+b²+c² = 2

Squaring both sides

=> a⁴+b⁴+c⁴ + 2(a²b²+ a²c²   +b²c²)  = 4

=> a⁴+b⁴+c⁴   = 4 - 2(a²b²+ a²c²   +b²c²)

=> a⁴+b⁴+c⁴   = 4 - 2(a²b²+ a²c²   +b²c²)

a+b+c = 1

Squaring both sides

=> a²+b²+c² + 2(ab + bc + ac)  = 1

=>  2 +  2(ab + bc + ac)  = 1

=> 2(ab + bc + ac)   = -1

=> ab + bc + ac = - 1/2

Squaring both sides

=> a²b²+ a²c²   +b²c²  + 2abc(a + b + c) = 1/4

=>  a²b²+ a²c²   +b²c²  + 2abc = 1/4

=> a²b²+ a²c²   +b²c²  = 1/4 - 2abc

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

=> 3 - 3abc  =  2 -  (-1/2)

=> 3 - 3abc  =  5/2

=>   3abc  =5/2

=> abc = 1/6

a²b²+ a²c²   +b²c²  = 1/4 - 2abc

=> a²b²+ a²c²   +b²c²  = 1/4 - 2/6

=> a²b²+ a²c²   +b²c²  =- 1/12

a⁴+b⁴+c⁴   = 4 - 2(a²b²+ a²c²   +b²c²)  

=>a⁴+b⁴+c⁴   = 4 - 2(-1/12)  

=> a⁴+b⁴+c⁴   =  25/6

=>  a⁴+b⁴+c⁴ =  4⅙​  

QED

Hence proved

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