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if x+y=3 and xy=-2 then (x^3+x+1)×(y^3+y+1)

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Answer to a math question if x+y=3 and xy=-2 then (x^3+x+1)×(y^3+y+1)

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Jayne
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(x^3+x+1)×(y^3+y+1)=x^3y^3+x^3y+xy^3+x^3+xy+x+y^3+y+1 So that: =(xy)^3+ xy(x^2+y^2)+xy+(x+y)+(x^3+y^3)+1 Now put xy=-2, and x+y=3: =(-2)^3+ (-2)(x^2+y^2)+(-2)+3+(x^3+y^3)+1 =-8-2(x^2+y^2)-2+3+(x^3+y^3)+1 =-6-2(x^2+y^2)+(x^3+y^3) Now to calculate x^2+y^2 use: (x+y)^2=3^2 x^2+y^2+2xy=9 x^2+y^2+2(-2)=9 x^2+y^2-4=9 x^2+y^2=13 and: x^3+y^3=(x+y)(x^2+y^2-xy)=3(13-(-2))=3(13+2)=3(15)=45 So that: =-6-2(x^2+y^2)+(x^3+y^3)=-6-2(13)+45 =-6-26+45 =-32+45 (x^3+x+1)×(y^3+y+1)=13

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