The number of all possible values of , where 0 < < , for which the system of equations (y + z) 3 = (xyz) 3 x 3 = 2 3 y + 2 3 z (xyz) 3 = (y + 2z) 3 + y 3 have a solution (x_0, y_0, z_0) with y_0 z_0 0, is

JEE Advanced 2010 Mathematics question

Not in JEE Main syllabus

The number of all possible values of , where , for which the system of equations

(y + z) 3θ = (xyz) 3θ

x 3θ = +

(xyz) 3θ = (y + 2z) 3θ + y 3θ

have a solution with , is

Answer: 3

Solution

Write , . Multiplying the second equation by gives . So the three equations give three expressions for : . From the first and third: , i.e. . From the first and

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Question ID VK-025788