If x(x - 3) = -1, then the value of x³(x³ - 18) is?

Explanation:-

$$\eqalign{ & x\left( {x - 3} \right) = - 1 \cr & \Rightarrow \left( {x - 3} \right) = \frac{{ - 1}}{x} \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x - 3} \right)^3} = {\left( {\frac{{ - 1}}{x}} \right)^3} \cr & \Rightarrow {x^3} - 27 - 9.x.\left( {x - 3} \right) = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 - 9 \times - 1 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 + 9 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 18 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3}\left( {{x^3} - 18} \right) = - 1 \cr}$$