Let $\tan { x } =a,\tan { y } =b,\tan { z } =c$ .

Given system is equal to

$a+b+c=6-\frac { 1 }{ a } -\frac { 1 }{ b } -\frac { 1 }{ c }$

${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=6-\frac { 1 }{ { a }^{ 2 } } -\frac { 1 }{ { b }^{ 2 } } -\frac { 1 }{ { c }^{ 2 } }$

${ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=6-\frac { 1 }{ { a }^{ 3 } } -\frac { 1 }{ { b }^{ 3 } } -\frac { 1 }{ { c }^{ 3 } }$ .

From second equation we complete squares to obtain

${ \left( a-\frac { 1 }{ a } \right) }^{ 2 }+{ \left( b-\frac { 1 }{ b } \right) }^{ 2 }+{ \left( c-\frac { 1 }{ c } \right) }^{ 2 }=0$

$\Rightarrow a=b=c=\pm 1$ .

Now rearranging 3rd equation and adding $3$ time first equation we get

${ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+\frac { 1 }{ { a }^{ 3 } } +\frac { 1 }{ { b }^{ 3 } } +\frac { 1 }{ { c }^{ 3 } } +3\left( a+\frac { 1 }{ a } +b+\frac { 1 }{ b } +c+\frac { 1 }{ c } \right) =6+18$

${ \left( a+\frac { 1 }{ a } \right) }^{ 3 }+{ \left( b+\frac { 1 }{ b } \right) }^{ 3 }+{ \left( c+\frac { 1 }{ c } \right) }^{ 3 }=24$

Therefore $\boxed {a=b=c=1}$ .

Make tan and cot trigonometric ratios with same #arguments in all 3 equations .

Sum and reciprocal of a no May have only 2 or as -2 as their sum.

Hence all tan's or cot's =+-1

But tanx=-1will not be satisfied

Hence tanx=1 answer indeed =6