已知x-y=1，x^3-y^3=2，求x^4+y^4及x^5-y^5的值
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（x³-y³)÷(x-y)=x²+xy+y²=2
(x-y)²=x²-2xy+y²=1 =>3(x²+y²)=5 =>x²+y²=5/3
=>3xy=1 => xy=1/3
x^4+y^4=(x²+x²)²-2x²y²
=25/9-2/9=23/9
x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
=(x^4+y^4)+xy(x^2+xy+y^2)
=23/9+(1/3)*2
=23/9+6/9
=29/9
(x-y)²=x²-2xy+y²=1 =>3(x²+y²)=5 =>x²+y²=5/3
=>3xy=1 => xy=1/3
x^4+y^4=(x²+x²)²-2x²y²
=25/9-2/9=23/9
x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
=(x^4+y^4)+xy(x^2+xy+y^2)
=23/9+(1/3)*2
=23/9+6/9
=29/9
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