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Answer:
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Explanation:
If a+b+c=1 , a²+b²+c²=2 , a³+b³+c³=3 then what is the value of a⁴+b⁴+c⁴?
If a+b+c=1, a2+b2+c2=2, a3+b3+c3=3 then what is the value of a4+b4+c4?
Using the identity,
a4+b4+c4=
(a2+b2+c2)22
+(4)(a+b+c)(a3+b3+c3)3
+(a+b+c)46
−(a+b+c)2(a2+b2+c2)
for this problem,
a4+b4+c4=(2)22+(4)(1)(3)3+(1)46−(1)2(2)=256
In case you wondered how I came up with that identity, I’ll tell you. No clever uses of roots of polynomials or clever mathematical tricks are needed, just brute force. I used Wolfram Alpha to expand
(w)(a2+b2+c2)2
+(x)(a+b+c)(a3+b3+c3)
+(y)(a+b+c)4
+(z)(a+b+c)2(a2+b2+c2)
and then I gathered together the terms like this:
(a4+b4+c4)(w+x+y+z)
+(a3b+a3c+b3a+b3c+c3a+c3b)(x+4y+2z)
+(a2bc+b2ac+c2ab)(12y+2z)
+(a2b2+b2c2+a2c2)(2w+6y+2z)
The coefficient of the first group of terms should be 1, and the coefficient of all the other terms should be zero, because I want the whole thing to add up to a4+b4+c4, so this gives me the following equations:
w+x+y+z=1
x+4y+2z=0
12y+2z=0
2w+6y+2z=0
Which then I solve using Cramer’s method, giving me w=12, x=43, y=16, and z=−1. The identity follows.
A variant of this approach will work to find any similar identity, if it exists.
