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 that,
\(2pq - 20 = 52 - 2pq\)
\(⇒ 4pq = 72\)
\(⇒ pq = 18\) ...... (1)

Now, \(p^2 + q^2 - 29 = 2pq - 20\)
\(⇒ p^2 + q^2 - 2pq = 9\)
\(⇒ (p- q)^2 =9\)
\(⇒ (p- q)^2 =\sqrt 9\)
\(⇒ p-q=±3\)

Also, \(p^2 + q^2 - 29 = 2pq - 20\)
\(⇒ p^2 + q^2 = 2pq + 9\)
\(⇒ p^2 + q^2 = 2(18) +9\)
\(⇒ p^2 + q^2 = 45\)

Now, \(p^3 - q^3 = (p—q) (p^2 + pq +q^2)\)
\(⇒ p^3 - q^3 = (p-q) (45 + 18)\)
\(⇒ p^3 - q^3 = (p - 9) (63)\)

If we put \(p-q = -3\)
\(p^3 - q^3 = 63(-3)\)
\(⇒ p^3 - q^3= -189\)

If we put \(p-q = 3\)
\(p^ - q^3 = 63(-3)\)
\(p^3 - q^3 = 189\)

The difference \(= 189 - (-189) = 189+189 = 378\)

So, the correct option is (B): \(378\).