Question

已知a+b+c=2,a^2+b^2+c^2=3,a^3+b^3+c^3=4,求a^4+b^4+c^4的值

速度一点啊!急求!

Answer 1

(a+b+c)^2=4
=a^2+b^2+c^2+2ab+2ac+2bc=3+2(ab+ac+bc)=4
ab+ac+bc=1/2
(a+b+c)^3
=a^3+b^3+c^3+3(ab+ac+bc)(a+b+c)-3abc
=4+3(1/2)(2)-3abc=4+3-3abc=8
abc=-1/3
(a+b+c)^4
=a^4+b^4+c^4+4a^3b+4a^3c+4b^3a+4b^3c+4c^3a+4c^3b+6a^2b^2+6a^c^2+6b^2c^2+12a^2bc+12ab^2c+12abc^2
=16
4(ab+ac+bc)(a^2+b^2+c^2)+3(a^2+b^2+c^2)(a^2+b^2+c^2)+8(a+b+c)abc-2(a^4+b^4+c^4)
=16=4*1/2(3)+3(3)(3)+8*2*(-1/3)-2(a^4+b^4+c^4)
16=12/2+27-16/3-2t
2t=17-16/3
2t=(51-16)/3=35/3
t=35/6

追问

答案是几😒😠

Answer 2

解：
a²+b²+c²=(a+b+c)²-2(ab+bc+ca)
ab+bc+ca=[(a+b+c)²-(a²+b²+c²)]/2=(2²-3)/2=1/2
a³+b³+c³=(a²+b²+c²)(a+b+c)-a²b-a²c-ab²-b²c-ac²-bc²
=3×2-(a²b+ab²)-(a²c+ac²)-(b²c+bc²)
=6-ab(a+b)-ac(a+c)-bc(b+c)
=6-ab(2-c)-ac(2-b)-bc(2-a)
=6-2ab+abc-2ac+abc-2bc+abc
=6-2(ab+bc+ca)+3abc
=6-1+3abc
=5+3abc=4
abc=-1/3

a⁴+b⁴+c⁴
=(a³+b³+c³)(a+b+c)-a³b-a³c-ab³-cb³-ac³-bc³
=4×2-(a³b+ab³)-(a³c+ac³)-(b³c+bc³)
=8-ab(a²+b²)-ac(a²+c²)-bc(b²+c²)
=8-ab(3-c²)-ac(3-b²)-bc(3-a²)
=8-3ab+abc²-3ac+ab²c-3bc+a²bc
=8-3(ab+bc+ca)+abc(a+b+c)
=8- 3/2 +(-1/3)×2
=8- 3/2 -2/3
=35/6