`p(q^(2)+r^(2))-q(r^(2)+p^(2))` can be factorised as

Options

A. `(P+q)(r^(2)-pq)`
B. `(p-q)(r^(2)-pq)`
C. `(p-q)(pq-r^(2))`
D. `(p-qr)(p+q)`

Answer

To factorise the expression \( p(q^2 + r^2) - q(r^2 + p^2) \), we will follow these steps:

### Step 1: Expand the expression
We start by expanding the expression:
\[ p(q^2 + r^2) - q(r^2 + p^2) = pq^2 + pr^2 - qr^2 - qp^2 \]

### Step 2: Rearrange the terms
Next, we rearrange the terms to group similar ones:
\[ pq^2 - qp^2 + pr^2 - qr^2 \]

### Step 3: Factor by grouping
Now we can factor by grouping. We will group the first two terms and the last two terms:
\[ (pq^2 - qp^2) + (pr^2 - qr^2) \]

### Step 4: Factor out common factors
From the first group \( pq^2 - qp^2 \), we can factor out \( pq \):
\[ pq(q - p) \]
From the second group \( pr^2 - qr^2 \), we can factor out \( r^2 \):
\[ r^2(p - q) \]

So, we rewrite the expression as:
\[ pq(q - p) + r^2(p - q) \]

### Step 5: Factor out the common binomial
Notice that both terms contain the factor \( (p - q) \):
\[ (p - q)(pq + r^2) \]

### Final Result
Thus, the expression \( p(q^2 + r^2) - q(r^2 + p^2) \) can be factorised as:
\[ (p - q)(pq + r^2) \]

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