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CAT 2023 Slot 2 QA Question & Solution

AlgebraMedium

Question

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

Options

243
486
378
189

Solution

Given that 2pq - 20 = 52 - 2pq => 4pq = 72 => pq = 18 ----(1)

Now, $p^2+q^2-29=2pq-20$ => $p^2+q^2-2pq=9$ => $\left(p-q\right)^2=9$ => $p-q=\pm\ 3$

Also, $p^2+q^2-29=2pq-20$ => $p^2+q^2=2pq+9=2\left(18\right)+9=45$

Now, $p^3-q^3=\left(p-q\right)\left(p^2+pq+q^2\right)$ = $\left(p-q\right)\left(45+18\right)$ = $\left(p-q\right)\left(63\right)$

=> When p-q = -3 => The value is 63(-3) = -189 and when p-q = 3 => The value is 63(3) = 189.

=> The difference = 189 - (-189) = 378.