Question

If \(p^2+q^2-29=2pq-20=52-2pq\) , then the difference between the maximum and minimum possible value of \((p^3-q^3 ) \)is

• A. 486
• B. 378
• C. 243
• D. 189

Answer 1

The Correct Option is B

Given:
\( 2pq - 20 = 52 - 2pq \) 

\( \Rightarrow 2pq + 2pq = 52 + 20 \)
\( \Rightarrow 4pq = 72 \)
\( \Rightarrow pq = 18 \)     ...... (1)

Now consider:

\( p^2 + q^2 - 29 = 2pq - 20 \)
\( \Rightarrow p^2 + q^2 - 2pq = 9 \)
\( \Rightarrow (p - q)^2 = 9 \)
\( \Rightarrow p - q = \pm 3 \)

From (1): \( pq = 18 \)

Again, using the same identity:
\( p^2 + q^2 = 2pq + 9 \)
\( \Rightarrow p^2 + q^2 = 2(18) + 9 = 36 + 9 = 45 \)

Now, using the identity:
\( p^3 - q^3 = (p - q)(p^2 + pq + q^2) \)

From above:
\( p^2 + q^2 = 45 \), and \( pq = 18 \)
So,
\( p^2 + pq + q^2 = 45 + 18 = 63 \)

Case 1: \( p - q = 3 \)

\( p^3 - q^3 = 3 \cdot 63 = 189 \)

Case 2: \( p - q = -3 \)

\( p^3 - q^3 = (-3) \cdot 63 = -189 \)

Difference between the two values:

\( 189 - (-189) = 189 + 189 = \mathbf{378} \)

Final Answer: (B) 378