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

Answer: (B)

378

Answer 2

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): $378$.