# what are x,y,z? y+x^3=3x(1+xy) z+y^3=3y(1+zy) x+z^3=3z(1+xz)

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Take the first equation and open the brackets.

`y+x^3=3x+3x^2*y`

`` Put the terms in y to the left, therefore subtract `3x^2*y. ` `y+x^3-3x^2*y=3x`

Put the terms in x only to the right, therefore subtract`x^3.` `y-3x^2*y=3x-x^3`

Factor, simultaneously, y to the left and x to the right.

`y(1-3x^2)=x(3-x^2)`

`` `y=[x(3-x^2)]/(1-3x^2)`

Put `x=tan alpha` and use formula for `tan 3 alpha` .

tan 3 alpha= `[tan alpha(3-tan^2 alpha)]/(1 - 3 tan^2 alpha)`

y=`tan 3 alpha`

Take the second equation and open the brackets.

`z+y^3=3y+3y^2*z`

Put the terms in z to the left, therefore subtract `3y^2*z` .

`z+y^3-3y^2*z=3y`

`z-3y^2*z=3y - y^3`

`` `z=[y(3-y^2)]/(1-3y^2)`

Since `y=tan 3 alpha =gt z = tan 3*(3 alpha) = tan 9 alpha` x = tan 3*(`9 alpha` ) = tan `27 alpha`

**Answer: The solution of the system is: tan alpha=tan 9 alpha = tan `27 alpha` **