Identity in Ramanujan style

Yes.

$$ \sqrt[3] {-1+7\sqrt[3]{20}} =\sqrt[3]{\frac{16}{9}}+\sqrt[3]{\frac{100}{9}}-\sqrt[3]{\frac{5}{9}}\tag0$$

Solution: More generally, given the three roots $x_i$ of any cubic equation,

$$x^3+ax^2+bx+c=0\tag1$$

then sums involving the cube roots of the $x_i$ can be given in the simple form,

$$(u+x_1)^{1/3}+(u+x_2)^{1/3}+(u+x_3)^{1/3} = \big(w+3\,\sqrt[\color{blue}6]{d}\big)^{1/3}$$

where $u,w$ are the constants,

$$u = \frac{ab-9c+\sqrt{d}}{2(a^2-3b)}\tag2$$

$$w = -\frac{(2a^3-9ab+27c)+9\sqrt{d}}{2(a^2-3b)}\tag3$$

and $d$ is,

$$d = \tfrac{1}{27}\Bigl(4(a^2-3b)^3-(2a^3-9ab+27c)^2\Bigr)\tag4$$

Example: For your question, we have,

$$w=-1, \quad d =\frac{7^6\cdot20^2}{3^6}$$

and subbing these into $(3),(4)$, and using $Mathematica$ to simplify, we get $b,c$ for arbitrary $a$ as,

$$b= \tfrac{1}{9}(-343+3a^2)$$

$$c= \tfrac{1}{27}(-2058-343a+a^3)$$

Substituting $b,c,d$ into $(2)$ and $(1)$, we find that,

$$u=\tfrac{1}{9}(37+3a)$$

and $(1)$ factors as,

$$ (7 + a + 3 x) (14 + a + 3 x) (-21 + a + 3 x) = 0\tag5$$

giving $x_1, x_2, x_3$. The expression $u+x_1$ simplifies as just $\frac{16}{9}$ and similarly for $x_2, x_3$, thus resolving into the numerical relation $(0)$ given above.