if a + b + c = 1, a^2 + b^2 + c^2 = 2, a^3+ b ^3+ c^3 = 3 then prove that
a^4+b^4+c^4= 25/6
Answers
Answer:
we know
2
(
a
b
+
b
c
+
c
a
)
=
(
a
+
b
+
c
)
2
−
(
a
2
+
b
2
+
c
2
)
⇒
2
(
a
b
+
b
c
+
c
a
)
=
1
2
−
2
=
−
1
⇒
a
b
+
b
c
+
c
a
=
−
1
2
given
a
3
+
b
3
+
c
3
=
3
⇒
a
3
+
b
3
+
c
3
−
3
a
b
c
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
c
a
)
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
(
a
b
+
b
c
+
c
a
)
+
3
a
b
c
=
3
⇒
(
1
×
(
2
−
(
−
1
2
)
+
3
a
b
c
)
)
=
3
⇒
(
2
+
1
2
)
+
3
a
b
c
=
3
⇒
3
a
b
c
=
3
−
5
2
=
1
2
⇒
a
b
c
=
1
6
Now
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
2
c
−
2
b
c
2
a
−
2
c
a
2
b
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
c
(
b
+
c
+
a
)
=
(
−
1
2
)
2
−
2
×
1
6
×
1
=
1
4
−
1
3
=
−
1
12
Now
a
4
+
b
4
+
c
4
=
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
2
2
−
2
×
(
−
1
12
)
=
4
+
1
6
=
4
1
6
Extension
a
5
+
b
5
+
c
5
=
(
a
3
+
b
3
+
c
3
)
(
a
2
+
b
2
+
c
2
)
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
c
2
)
]
=
3
⋅
2
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
]
Now
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
=
a
2
b
2
(
a
+
b
)
+
b
2
c
2
(
b
+
c
)
+
c
2
a
2
(
a
+
c
)
=
a
2
b
2
(
1
−
c
)
+
b
2
c
2
(
1
−
a
)
+
c
2
a
2
(
1
−
b
)
=
a
2
b
2
+
b
2
c
2
+
c
2
a
2
−
(
a
2
b
2
c
+
b
2
c
2
a
+
c
2
a
2
b
)
=
−
1
12
−
a
b
c
(
a
b
+
b
c
+
c
a
)
=
−
1
12
−
1
6
⋅
(
−
1
2
)
=
0
So
a
5
+
b
5
+
c
5
=
6
−
0
=
6
a+b+c=9 and a
2
+b
2
+c
2
=35
Using formula,
(a+b+c)
2
=a
2
+b
2
+c
2
+2(ab+bc+ca)
9
2
=35+2(ab+bc+ca)
2(ab+bc+ca)=81−35=46
(ab+bc+ca)=23
using formula,
(a
3
+b
3
+c
3
)−3abc=(a
2
+b
2
+c
2
−ab−bc−ca)(a+b+c)
a
3
+b
3
+c
3
−3abc=(35−23)×9=9×12=108
Answer (B) 108