Factorise : `(p-q)^(3)+(q-r)^(3)+(r-p)^(3)`

Answer

To factorize the expression \((p-q)^{3} + (q-r)^{3} + (r-p)^{3}\), we can use the identity for the sum of cubes. The identity states that:

\[ A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 + B^2 + C^2 - AB - BC - CA) \]

In our case, we can let:
- \(A = p - q\)
- \(B = q - r\)
- \(C = r - p\)

Now, we can observe that:

\[ A + B + C = (p - q) + (q - r) + (r - p) \]

Simplifying this, we have:

\[ A + B + C = p - q + q - r + r - p = 0 \]

Since \(A + B + C = 0\), we can apply the identity:

\[ A^3 + B^3 + C^3 = 3ABC \]

Now we can calculate \(ABC\):

\[ ABC = (p - q)(q - r)(r - p) \]

Thus, we have:

\[ (p - q)^{3} + (q - r)^{3} + (r - p)^{3} = 3(p - q)(q - r)(r - p) \]

So the final factorized form of the expression is:

\[ 3(p - q)(q - r)(r - p) \]

### Summary of Steps:
1. Identify \(A\), \(B\), and \(C\) as \(p - q\), \(q - r\), and \(r - p\).
2. Calculate \(A + B + C\) and show it equals zero.
3. Use the identity for the sum of cubes to express the original expression in terms of \(ABC\).
4. Calculate \(ABC\) and substitute back into the identity.
5. Write the final factorized form.