6 solutions in all

Algebra Level 3

Consider the system of equations

{ − 3 x y = − 7 2 − x 3 − y 3 = 20 27 \begin{cases} -3 xy = - \frac{7}{2} \\ -x^3 - y^3 = \frac{20}{27} \\ \end{cases}

There are 6 solutions ( x i , y i ) (x_i, y_i ) in total. If we let z 1 = x 1 + y 1 = x 2 + y 2 ; z 2 = x 3 + y 3 ; z 3 = x 4 + y 4 ; z 4 = x 5 + y 5 = x 6 + y 6 z_1=x_1+y_1=x_2+y_2 ; z_2=x_3+y_3 ; z_3=x_4+y_4 ; z_4=x_5+y_5=x_6+y_6 , then find ( 2 ∗ z 1 ) + z 2 + z 3 + ( 2 ∗ z 4 ) (2*z_1)+z_2+z_3+(2*z_4)


The answer is 0.

Raffaele Piccirillo by Raffaele Piccirillo , 28, Italy

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1 solution

Pulkit Gupta
Nov 25, 2015

Lets focus on the last equation.

( x + y ) 3 (x+y)^{3} = - 20/27

Now , ( x + y ) 3 (x+y)^{3} = ( x + y ) { ( x + y ) 2 (x+y)^{2} - 3 x y )

Putting x + y = z , we get a cubic equation in z with coefficient of ( z ) 2 (z)^{2} = 0 Hence the sum of all possible values of z is 0

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Yes,you found it :)

Raffaele Piccirillo - 5 years, 6 months ago

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