Given $x+\frac{1}{x}=5$, find $x^3+\frac{1}{x^3}$

The correct option is 110.

To find the value of $x^3+\frac{1}{x^3}$, we can use the algebraic identity for the cube of a sum:

${(a+b)}^3=a^3+b^3+3ab(a+b)$

Let $a=x$ and $b=\frac{1}{x}$. Substituting these into the identity gives:

${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3·x·\frac{1}{x}(x+\frac{1}{x})$

Simplifying the product term $x·\frac{1}{x}=1$, the equation becomes:

${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$

We are given that $x+\frac{1}{x}=5$. Substitute this value into the equation:

$5^3=x^3+\frac{1}{x^3}+3\left(5\right)$

Now, calculate the numerical values:
$5^3=125$
$3·5=15$

Substitute these values back into the equation:

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

Subtract 15 from both sides to isolate $x^3+\frac{1}{x^3}$:

$x^3+\frac{1}{x^3}=125-15$

$x^3+\frac{1}{x^3}=110$

Thus, the required value is 110.