Question

已知a+b=3,ab=-3,求a的五次方+b的五次方

Answer 1

a+b=3
a²+2ab+b²=9
a²+b²=15
a³+b³=(a+b)(a²-ab+b²)=54
(a²+b²)(a³+b³)
=a^5+b^5+a²b³+a³b²
=a^5+b^5+(a+b)a²b²
=a^5+b^5+27
a^5+b^5=(a²+b²)(a³+b³)-279=15*54-27=783

Answer 2

a+b=3
(a+b)^2=a^2+2ab+b^2=3^2=9
a^2+b^2=9-2ab=9-2*(-3)=15
a^3+b^3=(a+b)(a^2-ab+b^2)=3*[15-(-3)]=54
a^5+b^5
=(a^2+b^2)(a^3+b^3)-a^2b^3-a^3b^2
=15*54-a^2b^2(a+b)
=810-(ab)^2*(a+b)
=810-(-3)^2*3
=810-27=783

Answer 3

由已知可以得到a^2+b^2=(a+b)^2-2ab=15
a^5+b^5=(a+b)[(a^4+b^4)-ab(a^2+b^2)+a^2*b^2]
=(a+b)[(a^2+b^2)^2-2a^2*b^2-ab(a^2+b^2)+a^2*b^2]
=(a+b)[(a^2+b^2)^2-a^2*b^2-ab(a^2+b^2)]
=3*(15^2-3^2+3*15)
=783