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Algebra Level 4

If

{ x + y + z = 1 x 3 + y 3 + z 3 = 3 z 2 − 3 z + 1 x y z = 1 \begin{cases} x+y+z=1 \\ x^3 + y^3 + z^3 = 3z^2 -3z + 1 \\ \frac{xy}{z} = 1 \end{cases}

Find ∣ x + y z ∣ |\frac{x+y}{z}| .


The answer is 0.

Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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1 solution

Paul Ryan Longhas
Mar 28, 2016

Notice that x y z = 1 ⟹ z ≠ 0 , x y ≠ 0 \frac{xy}{z} = 1 \implies z \neq 0, xy \neq 0 .

Well, x + y + z = 1 ⟹ ( x + y ) 3 = ( 1 − z ) 3 = 1 − 3 z + 3 z 2 − z 3 = x 3 + y 3 x+y+z = 1 \implies (x+y)^3 = (1-z)^3 = 1 - 3z + 3z^2 - z^3 = x^3 + y^3

Thus, x 3 + 3 x 2 y + 3 x y 2 + y 3 = x 3 + y 3 ⟹ 3 x 2 y + 3 x y 2 = 0 x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 \implies 3x^2y + 3xy^2 = 0

So, 3 x y ( x + y ) = 0 3xy(x+y) = 0 . Since, x y ≠ 0 xy \neq 0 this force that x + y = 0 x+y = 0 .

Therefore, ∣ x + y z ∣ = 0 |\frac{x+y}{z}| = 0

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nicely done... (+1)..... I also found the solutions as x = ± i , y = ∓ i , z = 1 x=\pm i, y=\mp i, z=1 . It was not lengthy at all... Believe me... :-)

Rishabh Jain - 5 years, 2 months ago

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