Given three strictly positive real numbers $a, b, c$ and two strictly positive real numbers $d, e$ such that:

$$a^{2} + b^{2} + c^{2} = d^{2} + e^{2},$$
$$a^{4} + b^{4} + c^{4} = d^{4} + e^{4}.$$

We need to compare $a^{3} + b^{3} + c^{3}$ with $d^{3} + e^{3}$.

### Step 1: Use the Known Equalities

We have been given the equalities involving squares and fourth powers:

1. $a^{2} + b^{2} + c^{2} = d^{2} + e^{2}$ — This tells us that the sum of the squares of $a, b, c$ equal those of $d, e$.
2. $a^{4} + b^{4} + c^{4} = d^{4} + e^{4}$ — This similarly extends to the fourth powers.

### Step 2: Apply Power Mean Inequality

The Power Mean Inequality states that for any non-negative numbers $x_1, x_2, \ldots, x_n$ and any real numbers $p < q$,

$$\left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n} \right)^{1/p} \le \left( \frac{x_1^q + x_2^q + \cdots + x_n^q}{n} \right)^{1/q}.$$

#### Applying the Inequality:

- Let us apply this inequality to the sequences $a, b, c$ and $d, e$ with $p = 3$ and $q = 4$.
- For $a, b, c$:

$$\left( \frac{a^3 + b^3 + c^3}{3} \right)^{1/3} \le \left( \frac{a^4 + b^4 + c^4}{3} \right)^{1/4}.$$

- For $d, e$:

$$\left( \frac{d^3 + e^3}{2} \right)^{1/3} \le \left( \frac{d^4 + e^4}{2} \right)^{1/4}.$$

### Step 3: Analyzing Our Equations

Given $a^4 + b^4 + c^4 = d^4 + e^4$, the right sides of the inequalities are equal, which gives us:

$$\left( \frac{a^4 + b^4 + c^4}{3} \right)^{1/4} = \left( \frac{d^4 + e^4}{2} \right)^{1/4}.$$

Since $\frac{a^3 + b^3 + c^3}{3}$ measure follows the cube root mean of smaller numbers due to the positive nature and similar inequality holds for such values compared to fewer larger cubes in $d^3 + e^3$:

$$d^3 + e^3 \ge a^3 + b^3 + c^3.$$

Thus, the comparison yields:

$$\boxed{d^3 + e^3 \ge a^3 + b^3 + c^3}.$$