If x(x-3)=-1, then the value of x^{3}(x^{3}-1 | vedclass

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(C) Given equation: $x(x-3) = -1$$x^2 - 3x = -1$$x^2 - 3x + 1 = 0$We know that $(x^2 - 3x + 1) = 0$,so $x^2 + 1 = 3x$. Dividing by $x$,we get $x + \fr

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$-1$

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(C) Given equation: $x(x-3) = -1$$x^2 - 3x = -1$$x^2 - 3x + 1 = 0$We know that $(x^2 - 3x + 1) = 0$,so $x^2 + 1 = 3x$. Dividing by $x$,we get $x + \frac{1}{x} = 3$.We need to find the value of $x^3(x^3 - 18) = x^6 - 18x^3$.From $x + \frac{1}{x} = 3$,cubing both sides:$(x + \frac{1}{x})^3 = 3^3$$x^3 + \frac{1}{x^3} + 3(x)(\frac{1}{x})(x + \frac{1}{x}) = 27$$x^3 + \frac{1}{x^3} + 3(3) = 27$$x^3 + \frac{1}{x^3} = 27 - 9 = 18$$x^6 + 1 = 18x^3$$x^6 - 18x^3 = -1$Thus,the value of $x^3(x^3 - 18)$ is $-1$.

If $x(x-3)=-1,$ then the value of $x^{3}(x^{3}-18)$ is

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