MHB  Solve Quadratic System for $(2x-1)(2y-1)$

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For all real $a,\,b,\,x,\,y$ such that

$ax+by=4,\\ax^2+by^2=2,\\ax^3+by^3=-1.$

Find $(2x-1)(2y-1)$.
 
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anemone said:
For all real $a,\,b,\,x,\,y$ such that

$ax+by=4---(1),\\ax^2+by^2=2---(2),\\ax^3+by^3=-1---(3).$

Find $(2x-1)(2y-1)$.
$(2)\times x :ax^3+bxy^2=2x---(4)$
$(2)\times y :ax^2y+by^3=2y---(5)$
from $(1)(3)$ we get $(4)+(5)\rightarrow -1+4xy=2x+2y\rightarrow 4xy-2x-2y+1=1+1=2=(2x-1)(2y-1)$
 
Albert said:
$(2)\times x :ax^3+bxy^2=2x---(4)$
$(2)\times y :ax^2y+by^3=2y---(5)$
from $(1)(3)$ we get $(4)+(5)\rightarrow -1+4xy=2x+2y\rightarrow 4xy-2x-2y+1=1+1=2=(2x-1)(2y-1)$

Well done Albert, and thanks for participating!
 
Albert said:
$(2)\times x :ax^3+bxy^2=2x---(4)$
$(2)\times y :ax^2y+by^3=2y---(5)$
from $(1)(3)$ we get $(4)+(5)\rightarrow -1+4xy=2x+2y\rightarrow 4xy-2x-2y+1=1+1=2=(2x-1)(2y-1)$

Neat
 
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