If $$x + \frac{1}{x} = 5,  then  x^3 + \frac{1}{x^3}$$ is equal to:

As per given in the question,

$$x + \frac{1}{x} = 5$$

taking cube of both side,

$$\Rightarrow (x + \frac{1}{x} = 5)^3=x^3+\dfrac{1}{x^3}+3x\times{\dfrac{1}{x}}(x+\dfrac{1}{x})$$

Now, substituting the values in the question,

$$\Rightarrow (x + \frac{1}{x} = 5)^3=x^3+\dfrac{1}{x^3}+3x\times{\dfrac{1}{x}}(x+\dfrac{1}{x})$$

$$\Rightarrow (5^3)=x^3+\dfrac{1}{x^3}+3\times 5$$

$$\Rightarrow 125=x^3+\dfrac{1}{x^3}+15$$

$$\Rightarrow x^3+\dfrac{1}{x^3}=125-15=110$$