Determine the pair of constant numbers $ a,\ b,\ c$ such that for a quadratic function $ f(x) \equal{} x^2 \plus{} ax \plus{} b$, the following equation is identity with respect to $ x$.
\[ f(x \plus{} 1) \equal{} c\int_0^1 (3x^2 \plus{} 4xt)f'(t)dt\]
.
1. Given the quadratic function \( f(x) = x^2 + ax + b \), we first find its derivative:
   \[
   f'(x) = 2x + a
   \]

2. We need to evaluate the integral on the right-hand side of the given identity:
   \[
   c \int_0^1 (3x^2 + 4xt) f'(t) \, dt
   \]
   Substituting \( f'(t) = 2t + a \), we get:
   \[
   c \int_0^1 (3x^2 + 4xt)(2t + a) \, dt
   \]

3. Distribute and separate the integrals:
   \[
   c \left( \int_0^1 3x^2 (2t + a) \, dt + \int_0^1 4xt (2t + a) \, dt \right)
   \]

4. Evaluate each integral separately:
   \[
   \int_0^1 3x^2 (2t + a) \, dt = 3x^2 \left( \int_0^1 2t \, dt + \int_0^1 a \, dt \right)
   \]
   \[
   = 3x^2 \left( 2 \left[ \frac{t^2}{2} \right]_0^1 + a \left[ t \right]_0^1 \right)
   \]
   \[
   = 3x^2 (2 \cdot \frac{1}{2} + a \cdot 1) = 3x^2 (1 + a)
   \]

5. For the second integral:
   \[
   \int_0^1 4xt (2t + a) \, dt = 4x \left( \int_0^1 2t^2 \, dt + \int_0^1 at \, dt \right)
   \]
   \[
   = 4x \left( 2 \left[ \frac{t^3}{3} \right]_0^1 + a \left[ \frac{t^2}{2} \right]_0^1 \right)
   \]
   \[
   = 4x \left( 2 \cdot \frac{1}{3} + a \cdot \frac{1}{2} \right) = 4x \left( \frac{2}{3} + \frac{a}{2} \right)
   \]

6. Combine the results:
   \[
   c \left( 3x^2 (1 + a) + 4x \left( \frac{2}{3} + \frac{a}{2} \right) \right)
   \]

7. On the left-hand side, we have:
   \[
   f(x + 1) = (x + 1)^2 + a(x + 1) + b = x^2 + 2x + 1 + ax + a + b = x^2 + (a + 2)x + (a + b + 1)
   \]

8. Equate the coefficients from both sides:
   \[
   x^2 + (a + 2)x + (a + b + 1) \equiv c \left( 3x^2 (1 + a) + 4x \left( \frac{2}{3} + \frac{a}{2} \right) \right)
   \]

9. Matching the coefficients of \( x^2 \):
   \[
   1 = 3c(1 + a) \implies 3c(a + 1) = 1 \implies c = \frac{1}{3(a + 1)}
   \]

10. Matching the coefficients of \( x \):
    \[
    a + 2 = 4c \left( \frac{2}{3} + \frac{a}{2} \right)
    \]
    Substitute \( c = \frac{1}{3(a + 1)} \):
    \[
    a + 2 = 4 \cdot \frac{1}{3(a + 1)} \left( \frac{2}{3} + \frac{a}{2} \right)
    \]
    \[
    a + 2 = \frac{4}{3(a + 1)} \left( \frac{2}{3} + \frac{a}{2} \right)
    \]
    \[
    a + 2 = \frac{4}{3(a + 1)} \cdot \frac{4 + 3a}{6}
    \]
    \[
    a + 2 = \frac{4(4 + 3a)}{18(a + 1)}
    \]
    \[
    a + 2 = \frac{2(4 + 3a)}{9(a + 1)}
    \]
    \[
    9(a + 1)(a + 2) = 2(4 + 3a)
    \]
    \[
    9a^2 + 18a + 9 = 8 + 6a
    \]
    \[
    9a^2 + 12a + 1 = 0
    \]

11. Solve the quadratic equation:
    \[
    9a^2 + 12a + 1 = 0 \implies a = \frac{-12 \pm \sqrt{144 - 36}}{18} = \frac{-12 \pm 6\sqrt{3}}{18} = \frac{-2 \pm \sqrt{3}}{3}
    \]

12. For each \( a \), find \( b \) and \( c \):
    \[
    b = -(a + 1)
    \]
    \[
    c = \frac{1}{3(a + 1)}
    \]

    For \( a = -\frac{5}{3} \):
    \[
    b = \frac{2}{3}, \quad c = -\frac{1}{2}
    \]

    For \( a = -\frac{2}{3} \):
    \[
    b = -\frac{1}{3}, \quad c = 1
    \]

The final answer is \( \boxed{ \left( -\frac{5}{3}, \frac{2}{3}, -\frac{1}{2} \right), \left( -\frac{2}{3}, -\frac{1}{3}, 1 \right) } \)