What should be subtracted from $2 a^{2} - b^{2} + 3 a^{2} b$ to get $a^{2} + b^{2} + 3 a b^{2} - 6 ?$

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Solution

Let $p$ should be subtracted from $2 a^{2} - b^{2} + 3 a^{2} b$ to get $a^{2} + b^{2} + 3 a b^{2} - 6$ .

Therefore,
$2 a^{2} - b^{2} + 3 a^{2} b - p$ $= a^{2} + b^{2} + 3 a b^{2} - 6$
$p = 2 a^{2} - b^{2} + 3 a^{2} b - a^{2} - b^{2} - 3 a b^{2} + 6$
$p = a^{2} - 2 b^{2} + 3 a^{2} b - 3 a b^{2} + 6$

Hence, this is the answer.

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Question 58 (f)

Subtract:

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