Question

已知a+b+c=1，a^2+b^2+c^2=2，a^3+b^3+c^3=3，求abc的值。a^4+b^4+c^4

Answer 1

解：因为 (a+b+c)^2=a^2+b^2+c^2+2（ab+ac+bc）=1
=>ab+ac+bc= -1/2 .......@1

又有 (a+b+c)^3=3(a^2+b^2+c^2)(a+b+c)+6abc-2(a^3+b^3+c^3)
=>abc= 1/6 .......@2
由@1 =>(ab+ac+bc)^2=a^2*b^2+a^2*c^2+b^2*c^2+2abc(a+b+c)
=>a^2*b^2+a^2*c^2+b^2*c^2= (-1/2)^2-1/3= -1/12 ......@3
又因为 (a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2*b^2+a^2*c^2+b^2*c^2)
=>a^4+b^4+c^4=4-(-1/12)=25/6