已知a+b+c=0,求证:(a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3*(a^2+b^2+c^2)/2
1个回答

1/2(a^2+b^2+c^2)

=1/2[(a+b+c)^2-2(ab+bc+ca)]

=-(ab+bc+ca)

1/3(a^3+b^3+c^3)

=1/3[(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc]

=-abc

∴1/2(a^2+b^2+c^2)*1/3(a^3+b^3+c^3)=-abc(ab+bc+ca)

1/2(a^2+b^2+c^2)*1/3(a^3+b^3+c^3)

=1/6(a^5+a^2b^3+a^2c^3+a^3b^2+b^5+b^2c^3+a^3c^2+b^3c^2+c^5)

=1/6(a^5+b^5+c^5)+1/6[a^3(b^2+c^2)+b^3(a^2+c^2)+c^3(a^2+b^2)]

=1/6(a^5+b^5+c^5)+1/6{a^3[(b+c)^2-2bc]+b^3[(a+c)^2-2ac]+c^3[(a+b)^2-2ab]}

=1/6(a^5+b^5+c^5)+1/6{a^3[a^2-2bc]+b^3[b^2-2ac]+c^3[c^2-2ab]}

=1/6(a^5+b^5+c^5)+1/6{a^5-2a^3bc+b^5-2ab^3c+c^5-2abc^3}

=1/6(a^5+b^5+c^5)+1/6{(a^5+b^5+c^5)-2abc(a^2+b^2+c^2)}

=1/3(a^5+b^5+c^5)-1/3abc(a^2+b^2+c^2)

=1/3(a^5+b^5+c^5)-1/3abc[(a+b+c)^2-(2ab+2bc+2ca)]

=1/3(a^5+b^5+c^5)+2/3abc(ab+bc+ca)

∴1/3(a^5+b^5+c^5)+2/3abc(ab+bc+ca)=-abc(ab+bc+ca)

∴-abc(ab+bc+ca)=1/5(a^5+b^5+c^5)

∴1/2(a^2+b^2+c^2)*1/3(a^3+b^3+c^3)

=-abc(ab+bc+ca)

=1/5(a^5+b^5+c^5)