question from miscellaneous equations.

Question

if a+b+c=0, a³+b³+c³=3 and a⁵+b⁵+c⁵=10, then a⁴+b⁴+c⁴ is equal to

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Answer 1

We have given a+b+c=0, a³+b³+c³=3, a⁵+b⁵+c⁵=10,  a⁴+b⁴+c⁴=?
We know that (a+b+c)² = a²+b²+c²+2(ab+bc+ac)
=> 0 = a²+b²+c²+2(ab+bc+ac)

We have (a+b+c)(a²+b²+c²) = a³+b³+c³+a²b+b²a+b²c+c²b+c²a+a²c .........(1)
whereas,
(a+b+c)(ab+bc+ac) = a²b+b²a+b²c+c²b+c²a+a²c+3abc
a²b+b²a+b²c+c²b+c²a+a²c = -3abc .....(Since a+b+c = 0)................................(2)

 Now from eqation (1) and (2)
0 = a³+b³+c³-3abc
3abc = a³+b³+c³ =3   (since a³+b³+c³=3)
abc=1  -------------------(3)

Now
(a³+b³+c³)(a²+b²+c²)=a⁵+b⁵+c⁵+a³b²+b³c²+c³a²+a²b³+b²c³+c²a³ ---------------------(3)
a³b²+b³c²+c³a²+a²b³+b²c³+c²a³ = (a²b²+b²c²+c²a²)*(a+b+c)-(a²b²c+ab²c²+a²bc²)
= 0-(a²b²c+ab²c²+a²bc²)
= -3abc(ab+bc+ca)
= -3(ab+bc+ca) ......................(4)

Now form eqn. (3) and (4) we get
(a³+b³+c³)(a²+b²+c²)=a⁵+b⁵+c⁵-3(ab+bc+ca)
3(a²+b²+c²)=10 - 3(ab+bc+ca)
3(-2(ab+bc+ac))=10-3(ab+bc+ca)
ab+bc+ac=-10/3

also
3(a²+b²+c²)=10+3/2(a²+b²+c²)

a²+b²+c² = 20/3
(a²+b²+c²)²=(20/3)²
a⁴+b⁴+c⁴+2(a²b²+b²c²+c²a²)=400/9
(ab+bc+ca)²=a²b²+b²c²+c²a²+2(a²bc+ab²c+abc² )
a²b²+b²c²+c²a² =(-10/3)² - 2abc(a+b+c)=100/9
a⁴+b⁴+c⁴ = 400/9 - 2(100)/9=(400-200)/9=200/9

a⁴+b⁴+c⁴ = 200/9