Is it's possible!!!!!

From the expression: $3x^{ 2 }-15x+3=0\rightarrow x^{ 2 }-5x+1=0$ , therefore: $x^{2}+1=5x$

We need to find: $x^{3}+x^{-3}$ = $\left( x+\frac { 1 }{ x } \right) ^{ 3 }-3\left( x+\frac { 1 }{ x } \right)$ = $\left( \frac { x^{ 2 }+1 }{ x } \right) ^{ 3 }-3\left( \frac { x^{ 2 }+1 }{ x } \right)$

Using the expression above: $x^{ 3 }+\frac { 1 }{ x^{ 3 } } =\left( \frac { 5x }{ x } \right) ^{ 3 }-3\left( \frac { 5x }{ x } \right)$

x≠0: $x^{ 3 }+\frac { 1 }{ x^{ 3 } } =125-15=110$

Answer: $\boxed{110}$

dividing 3 from both sides and then cross multiplying, we get x^2 +6x+1=11x. so x^2 + 1 =5x...dividing both sides by x we get, x+ 1/x = 5...cubing both sides we get, x^3 + (1/x)^3 + 3(x+ 1/x)=125..now putting here x+ 1/x =5, we get x^3 + (1/x)^3 +15=125...so finally the answer is x^3 + (1/x)^3 =110...

3x^2 + 18x + 3 = 33x

3x^2 + 3 = 15x

x + 1/x = 5

x^3 + 1/x^3 = (x + 1/x)^3 - 3(x + 1/x) = 5^3 - 3(5) = 125 - 15 = 110