Foctorise :
` (1)/(3) x^(2) - (8)/(x)`

Answer

To factorise the expression \(\frac{1}{3} x^2 - \frac{8}{x}\), we will follow these steps:

### Step 1: Rewrite the expression with a common denominator
To combine the two terms, we need a common denominator. The common denominator for the terms \(\frac{1}{3} x^2\) and \(-\frac{8}{x}\) is \(3x\).

\[ \frac{1}{3} x^2 = \frac{x^3}{3x} \quad \text{and} \quad -\frac{8}{x} = -\frac{24}{3x} \]

So, we can rewrite the expression as:

\[ \frac{x^3 - 24}{3x} \]

### Step 2: Factor the numerator
Now, we need to factor the numerator \(x^3 - 24\). Notice that \(24\) can be expressed as \(2^3 \cdot 3\). Therefore, we can rewrite \(24\) as \(2^3\).

Now, we have:

\[ x^3 - (2\sqrt[3]{3})^3 \]

### Step 3: Apply the difference of cubes formula
We can use the difference of cubes formula, which states that:

\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

In our case, \(a = x\) and \(b = 2\sqrt[3]{3}\). Applying the formula gives us:

\[ x^3 - (2\sqrt[3]{3})^3 = (x - 2\sqrt[3]{3})(x^2 + 2\sqrt[3]{3}x + (2\sqrt[3]{3})^2) \]

### Step 4: Substitute back into the expression
Now substituting back into our expression, we have:

\[ \frac{(x - 2\sqrt[3]{3})(x^2 + 2\sqrt[3]{3}x + 4\cdot\sqrt[3]{3^2})}{3x} \]

### Final Answer
Thus, the factorised form of the expression \(\frac{1}{3} x^2 - \frac{8}{x}\) is:

\[ \frac{(x - 2\sqrt[3]{3})(x^2 + 2\sqrt[3]{3}x + 4\cdot\sqrt[3]{3^2})}{3x} \]