For $a\in \mathbb{R}$, consider the real valued function defined on $(-1,1)$ as follows:

For $x\neq 0$,

$f(x)=\frac{(1+x)^{\frac{1}{3}}-(1+2x)^{\frac{1}{4}}}{x}$

and for $x=0$,

$f(x)=a$

If $f$ is differentiable at $x=0$, then the value of $a+f'(0)$ is equal to:

Using expansion near $x=0$,

$(1+x)^{\frac{1}{3}}=1+\frac{x}{3}-\frac{x^2}{9}+O(x^3)$

Also,

$(1+2x)^{\frac{1}{4}}=1+\frac{x}{2}-\frac{3x^2}{8}+O(x^3)$

Now,

$(1+x)^{\frac{1}{3}}-(1+2x)^{\frac{1}{4}}$

$=\left(1+\frac{x}{3}-\frac{x^2}{9}\right)-\left(1+\frac{x}{2}-\frac{3x^2}{8}\right)+O(x^3)$

$=-\frac{x}{6}+\left(-\frac{1}{9}+\frac{3}{8}\right)x^2+O(x^3)$

$=-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)$

Therefore,

$f(x)=\frac{-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)}{x}$

$f(x)=-\frac{1}{6}+\frac{19x}{72}+O(x^2)$

For differentiability at $x=0$, function must be continuous at $x=0$.

So,

$a=\lim_{x\to 0}f(x)=-\frac{1}{6}$

Also,

$f'(0)=\frac{19}{72}$

Hence,

$a+f'(0)=-\frac{1}{6}+\frac{19}{72}$

$=-\frac{12}{72}+\frac{19}{72}$

$=\frac{7}{72}$