I've found the source of this interview.
It's in Paul Erdos' biography by Paul Hoffman, " The Man Who Love Only Numbers"(1998). Starting from page 78, the book describes Erdos left Hungary for Cambridge in 1934 due to the raging Hungarian Fascism. It was at Cambridge, the second day of his arrival, that he met G. H. Hardy, and inquired him about Ramanujan.
The quote "Erdos asked Hardy what his most important contribution
to mathematics was. 'The discovery of Ramanujan,' " appeared at the bottom of page 82.
If by "formula", Nirbhay (the former engineer featured in the article) means "closed form formula", then the answer is obviously not; because the answers would all have to be rational, whereas e.g the cube root of 2 is not. However, if by "formula" Nirbhay means "algorithm which computes the digits in sequence, which converges on the answer and terminates in the case of integers and other terminating fractions which are perfect cubes", then the answer is yes, if you allow either simple multiplications, or perform it in base 2. (Of course, you can reduce multiplication to addition if you like, but beyond small integers or shifting numbers by place values, that feels like cheating.)
No innovation is required; it's a simple modification of the digit-by-digit algorithm for computing square roots. The intuition behind this algorithm is that at each point, you are trying to get better and better approximations to X by constructing rational numbers Y1, Y2, Y3, ... having more and more digits,
whose cubes approximate X from below. We chip away at the error between X and the cubes of these approximations by trying to "complete the cube". Each new digit of X that we consider allows us to try to chip away more of the difference between Yj ³ and X, by defining Yj+1 = Yj + δ for a suitable increment δ; we then determine the error involved in the new approximation by computing Yj+1³ = Yj ³ + 3δYj ² + 3δ²Yj + δ³. The trick is how to do this with simple arithmetic operations; but it's not too difficult.
I demonstrate the algorithm in binary, because that's the base where it would be easiest in practise to compute the cube roots.
The generalization to arbitrary bases (e.g. decimal) is an exercise for the reader.
To simplify the presentation, we may assume without loss of generality that the number X whose cube-root we compute is less than 8. We may repeatedly divide by 8 until this is the case; when we have obtained the square root, simply multiply the answer the same number of times by 2. This is just observing that $\sqrt[3]{8^{-n} \cdot X\;} = 2^{-n} \cdot \sqrt[3]{X\;}$, and allows me to emphasize that the algorithm works just as well for fractions as for integers.
Because of how the algorithm works, it looks somewhat as though it's starting with a number in octal, and obtaining a root in binary (for square roots it looks like it produces a binary root from a number in quartal); but of course we can represent the base 8 "digits" by groups of three binary digits.
Let X = x0 + x1/81 + x2/82 + ... , where each xj ranges from 0 to 7. We compute binary digits yj representing Y = y0 + y1/21 + y2/22 + ... such that Y3 = X, as follows.
If x0 = 0, set y0 = 0; otherwise, set y0 = 1. Let R := x0 − y0.
Set A = B = C = y0 .
Set j = 1, and repeat the following until the desired number of digits is obtained (or until R = 0 and only if all the remaining digits xj , xj+1 , ... are also zero):
Set D = 8R + xj , and consider the largest binary digit yj such that
$$ D \;\;\geqslant\;\; y_j^3 + 6Ay_j^2 + 12By_j + 8C.$$
Of course, we're working with binary digits; all the powers of yj are either zero or one. So what we're really testing is whether or not
$$ D \;\;\geqslant\;\; 1 + 2A + 4A + 4B + 8B + 8C,$$
where multiplication by 2, 4, and 8 can be realized by shifting integers one, two, or three places (tacking on zeros). If the inequality above holds, set yj = 1; otherwise set yj = 0.
The integer A will actually be the number whose binary representation is the bit-sequence y0 y1 ... yj−1 in order; B is equal to A², and C is equal to A³. To maintain this invariant for the next iteration j, we compute new values A', B', and C' as updated values. For A, we define
$$ A' = 2A + y_j \;. $$
For B', we take advantage of the fact that we have B and A:
$$\begin{align*} B' &= (A')^2 = (2A + y_j)^2 = 4A^2 + 4Ay_j + y_j^2 \\&= 4B + 4Ay_j + y_j^2 \;. \end{align*}$$
Similarly, for C':
$$\begin{align*} C' &= (A')^3 = (2A + y_j)^2 = 8A^3 + 12A^2y_j + 6Ay_j^2 + y_j^3 \\&= 8C + 8By_j + 4By_j + 4Ay_j^2 + 2Ay_j^2 + y_j^3 \;. \end{align*}$$
Again, multiplications by zero, one, or powers of two are trivial inbinary representation. We then update A := A', B := B', and C := C'.
We update R, which is meant to represent the error of C as an approximation to the integer formed by the first j digits (in octal) of X: we set $$ R' = \bigl(8R + x_j\bigr) - C' = D - C' \;,$$ and then set R := R'.
If ever we obtain R = 0 and with all of the remaining digits of X equal to zero, we stop (with an exact solution).
In fact, it is not very difficult to modify this algorithm to obtain a procedure for extracting fourth roots, fifth roots, or nth roots for any constant n (though the number of parameters you have to carry around with you, and the sizes of the additions you have to carry out, will grow as you work harder and harder to avoid explicitly performing any multiplications).
Best Answer
When Ramanujan contacted Hardy he thought that he could provide a (nearly) exact formula for $\pi(n)$ the prime counting function. This formula implicitely supposed that $\zeta$ had no complex zero at all! This is neatly detailed by Hardy at the beginning of his book "Ramanujan". See too Berndt's "Ramanujan and the theory of prime numbers" for some details.