Find the value of $\sqrt[3]{26}$ using Mean Value Theorem
My Try:
We choose the function $f(x)=\sqrt[3]{x}$, in the interval $x \in [26,27]$. So for $0<\theta<1$, we have $$\frac{f(27)-f(26)}{27-26}=f'(26+\theta) \Rightarrow3-f(26)=\frac{1}{3}\left(26+\theta\right)^{\frac{-2}{3}}$$
Now since $\theta<1$, we have $$3-f(26)>\frac{1}{3}27^{\frac{-2}{3}}=\frac{1}{27} $$
$$\Rightarrow f(26)<\frac{80}{27}$$
Now I am not sure what do to next?