The interval on which the function $f(x) = x^4 - \frac{x^3}{3}$ is strictly decreasing, is:

A function is strictly decreasing on an interval when the derivative is negative.

Given: $f(x) = x^4 - \frac{x^3}{3}$

Finding the derivative using the power rule:

$f'(x) = 4x^3 - \frac{3x^2}{3}$

$f'(x) = 4x^3 - x^2$

---

Factoring out common terms:

$f'(x) = x^2(4x - 1)$

---

Setting $f'(x) = 0$:

$x^2(4x - 1) = 0$

This gives:

$x^2 = 0$ → $x = 0$

$4x - 1 = 0$ → $x = \frac{1}{4}$

Critical points: $x = 0$ and $x = \frac{1}{4}$

---

Analyzing the sign of $f'(x) = x^2(4x - 1)$ in each interval:

Note that $x^2$ is always non-negative. The sign of $f'(x)$ depends on $(4x - 1)$.

For $x < \frac{1}{4}$:

$(4x - 1) < 0$, so $f'(x) \leq 0$

The function is decreasing.

For $x > \frac{1}{4}$:

$(4x - 1) > 0$, so $f'(x) > 0$

The function is increasing.

---

The function is strictly decreasing on $\left(-\infty, \frac{1}{4}\right)$.