Its 125-15=110

We have $x+\frac{1}{x}=5\implies \frac{x^2+1}{x}=5\implies x^2+1=5x\implies x^2-5x+1=0$. Using the quadratic formula we have $x=\frac{5}{2}\pm\frac{\sqrt{21}}{2}$. Plugging either value into $x^3 +\frac{1}{x^3} $ we get our final answer, $x^3 +\frac{1}{x^3} =\boxed{110}$.

$(x+\frac1x)^2 = x^2 + 2 + \frac1{x^2} = 25$ $(x^2 + \frac1{x^2})(x+\frac1x) = x^3 + \frac1{x^3} + x+\frac1x $ $x^3+\frac1{x^3} = 23 \cdot 5 - 5 =110$

$$(x+\frac{1}{x})^3=x^3+3 x^2\cdot \frac{1}{x}+3 x \cdot \frac{1}{x^2}+\frac{1}{x^3} \Rightarrow $$ $$x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})=(5)^3-3(5)=110$$

Pretty common type problem. As a general rule, when given $x + \frac{1}{x}$ and wanting to find $x^n + \frac{1}{x^n}$, you can use the recursive relationship, find all $x^y + \frac{1}{x^y}$ where $y$ is between $1$ and $x$ and then substitute.

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Squaring gives, $$x^2+\frac{1}{x^2}+2=25 \implies x^2+\frac{1}{x^2}=23$$Multiplying, $$\left(x^2+\frac{1}{x^2} \right) \left(x+\frac{1}{x}\right)=x^3+x+\frac{1}{x}+\frac{1}{x^3}=115 \implies x^3+\frac{1}{x^3}=\boxed{110}$$

This is very similar to the Chebyshev polynomials. If we let $\mathcal{T}_0(x) = 2, \mathcal{T}_1(x) = x, \mathcal{T}_{n+1}(x) = x\mathcal{T}_n - \mathcal{T}_{n-1}$ then these polynomials satisfy $$\mathcal{T}_{n}\left(t+\frac1t\right) = t^n + \frac1{t^n}$$and also $T_n(x) = \frac12\mathcal{T}_n(2x)$ where $T_n$ is the $n$th Chebyshev polynomial of the first kind. Source: Putnam and Beyond by Gelca and Andreescu

We can simply use the formula of (a+b)^3 (x+1/x)^3 = x^3 + (1/x)^3+ 3(x×(1/x)) (x+(1/x)) 125= x^3 + (1/x)^3 +3×5 Therefore x^3+(1/x)^3=125-15 =110(answer)

Cubing the first equation, $$(x+\frac{1}{x})^3=125$$$$x^3+3(x+\frac{1}{x})+\frac{1}{x^3}=125$$$$x^3+\frac{1}{x^3}=125-3(5)=110$$

$(x+\frac{1}{x})^3=5^3$ $x^3+3x+\frac{3}{x}+\frac{1}{x^3} = 125$ $x^3 + 3(x+\frac{1}{x}) + \frac{1}{x^3} = 125$ $x^3+\frac{1}{x^3} = \boxed{110}$

$x+\dfrac{1}{x}=5$. $(x+\dfrac{1}{x})^3=5^3$. Note that $x^3+3x+\dfrac{3}{x}+\dfrac{1}{x^3}=125$. In order to find the value of $x^3+\dfrac{1}{x^3}$, we must isolate them both. $x^3+3(x+\dfrac{1}{x})+\dfrac{1}{x^3}=125$. $x+\dfrac{1}{x}=5$ $x^3+\dfrac{1}{x^3}+15=125$ $x^3+\dfrac{1}{x^3}=\boxed{110}$