Question

已知:x+y=3,x^2+y^2+xy=8,则x^4+y^4+3x^4y+3xy^4的ŀ

Answer 1

x+y=3,
x^2+y^2+xy=(x+y)^2-xy=9-xy=8,
xy=1,
所以x^2+y^2=(x+y)^2-2xy=7,
x^3+y^3=(x+y)(x^2-xy+y^2)=3*6=18,
x^4+y^4=(x^2+y^2)^2-2(xy)^2=49-2=47,
于是x^4+y^4+3x^4y+3xy^4
=47+3xy(x^3+y^3)
=47+3*18
=47+54
=101.

Answer 2

(x+y)²-(x²+y²+xy)
=(x²+2xy+y²)-(x²+y²+xy)
=xy=3²-8=1

x+y=3 xy=1

x^4+y^4+3x^4y+3xy^4
=x^4+y^4+3xy(x^3+y^3)
=(x²+y²)²-2x²y² +3xy(x+y)(x²-xy+y²)
=[(x+y)²-2xy]²-2x²y²+3xy(x+y)(x²+y²+xy-2xy)
=(3²-2)²-2×1²+3×1×3×(8-2×1)
=101