To identify the intervals where \( f(x) \) is strictly decreasing, we examine its derivative, \( f'(x) \).
The derivative of \( f(x) = 2x^3 - 15x^2 - 144x - 7 \) is calculated as follows:
\[
f'(x) = \frac{d}{dx}(2x^3 - 15x^2 - 144x - 7).
\]
Step 1: Compute \( f'(x) \).
Differentiating term by term yields:
\[
f'(x) = 6x^2 - 30x - 144.
\]
Step 2: Find the critical points by setting \( f'(x) = 0 \).
We factorize \( f'(x) \):
\[
6x^2 - 30x - 144 = 0.
\]
Dividing the equation by 6 simplifies it to:
\[
x^2 - 5x - 24 = 0.
\]
Factoring the quadratic expression gives:
\[
(x - 8)(x + 3) = 0.
\]
The critical points are therefore:
\[
x = -3, \quad x = 8.
\]
Step 3: Analyze the sign of \( f'(x) \) in the intervals defined by the critical points.
The critical points \( x = -3 \) and \( x = 8 \) partition the real number line into three intervals: \( (-\infty, -3) \), \( (-3, 8) \), and \( (8, \infty) \). We test a point within each interval to determine the sign of \( f'(x) \).
- In the interval \( (-\infty, -3) \), let's choose \( x = -4 \).
\[
f'(-4) = 6(-4)^2 - 30(-4) - 144 = 96 + 120 - 144 = 72.
\]
Since \( f'(-4) = 72>0 \), \( f(x) \) is increasing in this interval.
- In the interval \( (-3, 8) \), let's choose \( x = 0 \).
\[
f'(0) = 6(0)^2 - 30(0) - 144 = -144.
\]
Since \( f'(0) = -144<0 \), \( f(x) \) is decreasing in this interval.
- In the interval \( (8, \infty) \), let's choose \( x = 9 \).
\[
f'(9) = 6(9)^2 - 30(9) - 144 = 486 - 270 - 144 = 72.
\]
Since \( f'(9) = 72>0 \), \( f(x) \) is increasing in this interval.
Step 4: Conclude the interval of strict decrease.
Based on the sign analysis of \( f'(x) \), \( f(x) \) is strictly decreasing on the interval \( (-3, 8) \).
Final Answer:
\[
\boxed{(-3, 8)}
\]