Algebra Anyone?

Algebra Level 4

Let x x , y y , and z z be non-zero real numbers such that x y + y z + z x = 0 xy + yz + zx = 0 and ( x + y + z + 1 ) 2 = x y z (x + y + z + 1)^2 = xyz .

Determine the maximum value of ( x y − z ) ( z x − y ) ( y z − x ) (xy - z)(zx - y)(yz - x) .


The answer is 0.

Valentio Iverson by Valentio Iverson , 17, Indonesia

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1 solution

P = ( x y − z ) ( z x − y ) ( y z − x ) Since x y + y z + z x = 0 = ( − y z − z x − z ) ( − x y − y z − y ) ( − x y − z x − x ) = − x y z ( x + y + 1 ) ( z + x + 1 ) ( y + z + 1 ) = − x y z ( x + y + z + 1 − z ) ( x + y + z + 1 − y ) ( x + y + z + 1 − x ) As ( x + y + z + 1 ) = x y z = − x y z ( x y z − z ) ( x y z − y ) ( x y z − x ) Let a = x y z = − a 2 ( a − z ) ( a − y ) ( a − x ) = − a 2 ( a 3 − ( x + y + z ) a 2 + ( x y + y z + z x ) a − x y z ) Since x y + y z + z x = 0 = − a 2 ( a 3 − ( x + y + z + 1 − 1 ) a 2 + 0 − a 2 ) Note that x + y + z + 1 = a = − a 2 ( a 3 − ( a − 1 ) a 2 − a 2 ) = − a 2 ( a 3 − a 3 + a 2 − a 2 ) = 0 \begin{aligned} P & = ({\color{#3D99F6}xy}-z)({\color{#3D99F6}zx}-y)({\color{#3D99F6}yz}-x) & \small \color{#3D99F6} \text{Since }xy+yz+zx = 0 \\ & = ({\color{#3D99F6}-yz-zx}-z)({\color{#3D99F6}-xy-yz}-y)({\color{#3D99F6}-xy-zx}-x) \\ & = -xyz(x+y+1)(z+x+1)(y+z+1) \\ & = -xyz({\color{#3D99F6}x+y+z+1}-z)({\color{#3D99F6}x+y+z+1}-y)({\color{#3D99F6}x+y+z+1}-x) & \small \color{#3D99F6} \text{As }(x+y+z+1) = xyz \\ & = -xyz({\color{#3D99F6}\sqrt{xyz}}-z)({\color{#3D99F6}\sqrt{xyz}}-y)({\color{#3D99F6}\sqrt{xyz}}-x) & \small \color{#3D99F6} \text{Let } a = \sqrt{xyz} \\ & = -a^2(a-z)(a-y)(a-x) \\ & = -a^2\left(a^3-(x+y+z)a^2+{\color{#3D99F6}(xy+yz+zx)}a -xyz \right) & \small \color{#3D99F6} \text{Since }xy+yz+zx = 0 \\ & = -a^2\left(a^3-({\color{#D61F06}x+y+z+1}-1)a^2+{\color{#3D99F6}0} -a^2 \right) & \small \color{#D61F06} \text{Note that }x+y+z+1 = a \\ & = -a^2\left(a^3-({\color{#D61F06}a}-1)a^2-a^2 \right) \\ & = -a^2\left(a^3-a^3+a^2-a^2 \right) \\ & = \boxed 0 \end{aligned}

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