A strange algebra question

Algebra Level 3

Given x , y ∈ R x,y \in \mathbb{R} that x y − x y 2 − y x 2 = 3 xy-\dfrac{x}{y^2}-\dfrac{y}{x^2}=3 . Find the sum of all possible value of ( x − 1 ) ( y − 1 ) (x-1)(y-1) .


The answer is 5.000.

Isaac Yiu Math Studio by Isaac YIU Math Studio , 15, Hong Kong

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1 solution

Multiply by x 2 y 2 x^2y^2 both sides and rearrange: x 3 y 3 − x 3 − y 3 − 3 x 2 y 2 = 0 x^3y^3-x^3-y^3-3x^2y^2=0 Let a = x y , b = − x , c = − y a=xy,b=-x,c=-y . Then: a 3 + b 3 + c 3 − 3 a b c = 0 ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) = 0 a + b + c = 0 ∪ a 2 + b 2 + c 2 − a b − b c − c a = 0 a^3+b^3+c^3-3abc=0 \\ (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0 \\ a+b+c=0 \cup a^2+b^2+c^2-ab-bc-ca=0 Case 1: a + b + c = 0 a+b+c=0 a + b + c = 0 x y − x − y = 0 x y − x − y + 1 = 1 ( x − 1 ) ( y − 1 ) = 1 a+b+c=0 \\ xy-x-y=0 \\ xy-x-y+1=1 \\ (x-1)(y-1)=1 Case 2: a 2 + b 2 + c 2 − a b − b c − c a = 0 a^2+b^2+c^2-ab-bc-ca=0 a 2 + b 2 + c 2 − a b − b c − c a = 0 1 2 [ ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 ] = 0 a = b = c x y = − x = − y x = y = − 1 ( x − 1 ) ( y − 1 ) = ( − 1 − 1 ) 2 = 4 a^2+b^2+c^2-ab-bc-ca=0 \\ \dfrac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0 \\ a=b=c \\ xy=-x=-y \\ x=y=-1 \\ (x-1)(y-1)=(-1-1)^2=4 Therefore, the sum is 1 + 4 = 5 1+4=\boxed{5}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

The equations in Case 2 give the case x = y = 0 x=y=0 as well, but that case can be discounted as not satisfying the original equation.

Mark Hennings - 1 year, 7 months ago

Yes, you are right, so I didn't say it.

Isaac YIU Math Studio - 1 year, 7 months ago

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