Question

If a+b+c=1
a^2+b^2+c^2=2 and
a^3+b^3+c^3=3 then find the value of
a^4+b^4+c^4 ?

Answer 1

Given ,
a + b + c = 1
a² + b² + c² = 2
a³ + b³ + c³ = 3

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
1² = 2 + 2(ab + bc + ca)
-1/2 = (ab + bc + ca)-------(1)

again,
a³ + b³ + c³ - 3abc = (a + b + c){a² + b² + c² -(ab + bc + ca)}
3 - 3abc = (1) { 2 - (-1/2) }
3 - 3abc = 5/2
-3abc = -1/2
abc = 1/6 -----(2)

now,
a⁴ + b⁴ + c⁴ = {a² + b² + c² }² - 2{a²b² + b²c² + c²a² }
= (2)² - 2{(ab + bc + ca) ² - 2(ab.bc + bc.ca + ca.ab)}
= 4 - 2{ (-1/2)² - 2abc(a + b +c)}
= 4 - 2{ 1/4 -2 × 1/6 × (-1/2)}
= 4 - 2{ 1/4 + 1/6 }
= 4 - 5/6
= 19/6

Answer 2

Given : a+b+c = 1

a²+b²+c² = 2

a³+b³+c³ = 3

To Find : a⁴+b⁴+c⁴

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

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