If $$[t]$$ denotes the greatest integer $$\leq 1$$, then the value of $$\frac{3(e-1)}{e} \int_1^2 x^2 e^{[x]+[x^3]} dx$$ is:

Step 1: Simplify the greatest integer part $$[x]$$

The interval of integration is from $$x = 1$$ to $$x = 2$$. For any value of $$x$$ in the open interval $$(1, 2)$$, the greatest integer function $$[x]$$ yields a constant value:

$$[x] = 1$$

Substituting $$[x] = 1$$ into the expression simplifies the integral to:

$$I = \int_1^2 x^2 e^{1 + [x^3]} dx = e \int_1^2 x^2 e^{[x^3]} dx$$

Step 2: Use substitution to transform the limits and the integrand

Let $$u = x^3$$. Differentiating both sides gives:

$$du = 3x^2 dx \implies x^2 dx = \frac{1}{3} du$$

Now we find the corresponding limits for $$u$$:

- When $$x = 1$$, $$u = 1^3 = 1$$

- When $$x = 2$$, $$u = 2^3 = 8$$

Substituting these components back into the integral gives:

$$I = e \int_1^8 e^{[u]} \cdot \frac{1}{3} du = \frac{e}{3} \int_1^8 e^{[u]} du$$

Step 3: Evaluate the integral by breaking it into integer subintervals

The value of $$[u]$$ remains constant within each unit step from $$u = 1$$ to $$u = 8$$. We expand the integral over these discrete intervals:

$$\int_1^8 e^{[u]} du = \int_1^2 e^1 du + \int_2^3 e^2 du + \int_3^4 e^3 du + \dots + \int_7^8 e^7 du$$

$$\int_1^8 e^{[u]} du = e^1(1) + e^2(1) + e^3(1) + e^4(1) + e^5(1) + e^6(1) + e^7(1)$$

This forms a geometric progression with first term $$a = e$$, common ratio $$r = e$$, and $$n = 7$$ terms:

$$\sum_{k=1}^7 e^k = \frac{e(e^7 - 1)}{e - 1}$$

Step 4: Combine the parts to find the final value

Substitute the summation back into the equation for $$I$$:

$$I = \frac{e}{3} \cdot \left( \frac{e(e^7 - 1)}{e - 1} \right) = \frac{e^2(e^7 - 1)}{3(e - 1)}$$

Now, multiply by the external factor given outside the integral operator:

$$\text{Value} = \frac{3(e-1)}{e} \cdot I$$

$$\text{Value} = \frac{3(e-1)}{e} \cdot \frac{e^2(e^7 - 1)}{3(e - 1)}$$

$$\text{Value} = e(e^7 - 1) = e^8 - e$$

Conclusion:

The value of the expression is equal to $$e^8 - e$$.