Determine all real numbers $a, b, c, d$ such that the polynomial $f(x) = ax^3 +bx^2 + cx + d$ satisfies simultaneously the folloving conditions $\begin {cases} |f(x)| \le 1 \,for \,  |x| \le 1 \\ f(2) = 26 \end {cases}$
To determine all real numbers \(a, b, c, d\) such that the polynomial \(f(x) = ax^3 + bx^2 + cx + d\) satisfies the conditions:
\[
\begin{cases}
|f(x)| \le 1 \quad \text{for} \quad |x| \le 1 \\
f(2) = 26
\end{cases}
\]

1. **Using the given conditions:**
   - We know that \(|f(x)| \le 1\) for \(|x| \le 1\).
   - We also know that \(f(2) = 26\).

2. **Express \(f(2)\) using the polynomial:**
   \[
   f(2) = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d
   \]
   Given \(f(2) = 26\), we have:
   \[
   8a + 4b + 2c + d = 26 \quad \text{(Equation 1)}
   \]

3. **Using Chebyshev polynomials:**
   - The Chebyshev polynomials of the first kind \(T_n(x)\) are known to satisfy \(|T_n(x)| \le 1\) for \(|x| \le 1\).
   - For \(T_3(x) = 4x^3 - 3x\), we have \(|T_3(x)| \le 1\) for \(|x| \le 1\).

4. **Constructing \(f(x)\) using Chebyshev polynomials:**
   - We can try \(f(x) = kT_3(x)\) where \(T_3(x) = 4x^3 - 3x\).
   - Thus, \(f(x) = k(4x^3 - 3x)\).

5. **Determine \(k\) such that \(f(2) = 26\):**
   \[
   f(2) = k(4(2)^3 - 3(2)) = k(32 - 6) = 26
   \]
   \[
   26k = 26 \implies k = 1
   \]

6. **Thus, the polynomial \(f(x)\) is:**
   \[
   f(x) = 4x^3 - 3x
   \]

7. **Comparing with the general form \(f(x) = ax^3 + bx^2 + cx + d\):**
   - We have \(a = 4\), \(b = 0\), \(c = -3\), and \(d = 0\).

8. **Verification:**
   - Check \(|f(x)| \le 1\) for \(|x| \le 1\):
     \[
     f(x) = 4x^3 - 3x
     \]
     For \(|x| \le 1\), \(|4x^3 - 3x| \le 1\) holds true.
   - Check \(f(2) = 26\):
     \[
     f(2) = 4(2)^3 - 3(2) = 32 - 6 = 26
     \]

Thus, the only solution is \((a, b, c, d) = (4, 0, -3, 0)\).

The final answer is \( \boxed{ (4, 0, -3, 0) } \).