Answer :
Step-by-step explanation:
The given equation is:
\[ p^3 + q^3 + r^3 - 3pqr = \frac{1}{2}(p+q+r)\left[ (p-q)^2 + (q-r)^2 + (r-p)^2 \right] \]
This equation is a specific case of the identity related to the sum and products of roots of a polynomial, and it is true under certain conditions, typically when \( p, q, r \) are roots of a cubic polynomial. Let's verify it step by step.
Firstly, recall the identity for the sum of cubes:
\[ p^3 + q^3 + r^3 - 3pqr = (p+q+r)(p^2 + q^2 + r^2 - pq - qr - rp) \]
Now, let's simplify the right-hand side of the given equation:
\[ \frac{1}{2}(p+q+r)\left[ (p-q)^2 + (q-r)^2 + (r-p)^2 \right] \]
Expand and simplify each squared term:
\[ (p-q)^2 = p^2 - 2pq + q^2 \]
\[ (q-r)^2 = q^2 - 2qr + r^2 \]
\[ (r-p)^2 = r^2 - 2rp + p^2 \]
Add these together:
\[ (p-q)^2 + (q-r)^2 + (r-p)^2 = p^2 + q^2 + r^2 - 2(pq + qr + rp) \]
Multiply by \( \frac{1}{2}(p+q+r) \):
\[ \frac{1}{2}(p+q+r)(p^2 + q^2 + r^2 - 2(pq + qr + rp)) \]
This expression matches the right-hand side of the original equation. Therefore, the given equation holds true:
\[ p^3 + q^3 + r^3 - 3pqr = \frac{1}{2}(p+q+r)\left[ (p-q)^2 + (q-r)^2 + (r-p)^2 \right] \]
This identity is useful in algebraic manipulations involving sums and products of roots of polynomials, particularly cubic ones.