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

Options

A. `2pq(p+q)(q+r)(r-p)`
B. `3pqr(p-q)(r-q)(r-p)`
C. `2pqr(p-q)(q-r)(p-r)`
D. `3pqr(p-q)(q-r)(r-p)`

Answer

To factorise the expression \( p^3(q - r)^3 + q^3(r - p)^3 + r^3(p - q)^3 \), we can use the identity for the sum of cubes. The identity states that if \( a + b + c = 0 \), then:

\[ a^3 + b^3 + c^3 = 3abc \]

### Step-by-Step Solution:

1. **Identify \( a, b, c \)**:
- Let \( a = p(q - r) \)
- Let \( b = q(r - p) \)
- Let \( c = r(p - q) \)

2. **Check if \( a + b + c = 0 \)**:
\[ a + b + c = p(q - r) + q(r - p) + r(p - q) \]
Expanding this:
\[ = pq - pr + qr - qp + rp - rq \]
Rearranging the terms:
\[ = (pq - qp) + (qr - rq) + (rp - pr) = 0 \]
Since \( a + b + c = 0 \), we can apply the identity.

3. **Apply the identity**:
According to the identity:
\[ a^3 + b^3 + c^3 = 3abc \]
Therefore,
\[ p^3(q - r)^3 + q^3(r - p)^3 + r^3(p - q)^3 = 3 \cdot p(q - r) \cdot q(r - p) \cdot r(p - q) \]

4. **Calculate \( abc \)**:
\[ abc = p(q - r) \cdot q(r - p) \cdot r(p - q) \]

5. **Final Factorization**:
Thus, the expression can be factored as:
\[ = 3pqr(q - r)(r - p)(p - q) \]

### Final Answer:
The factorized form of the expression \( p^3(q - r)^3 + q^3(r - p)^3 + r^3(p - q)^3 \) is:
\[ 3pqr(q - r)(r - p)(p - q) \]