Add $(6 a b^{3} - 5 a b) (2 a^{2} + b - 5), a b (a^{2} + 1)$ and $a b^{3} (3 a^{3} + 2 b + 1) .$

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Solution

Given that,

$(6 a b^{3} - 5 a b) (2 a^{2} + b - 5) + a b (a^{2} + 1) + a b^{3} (3 a^{3} + 2 b + 1)$

$= 12 a^{3} b^{3} + 6 a b^{4} - 30 a b^{3} - 10 a^{2} b - 5 a b^{2} - 25 a b + a^{3} b + a b + 3 a^{4} b^{3} + 2 a b^{4} + a b^{3}$

$= 12 a^{3} b^{3} + 8 a b^{4} - 29 a b^{3} - 10 a^{2} b - 5 a b^{2} - 2$ $a b + a^{3} b + 3 a^{4} b^{3}$

$= 3 a^{3} b^{3} (4 + a) - 5 a b (2 a + b) + a b^{3} (8 b - 29) + a b (a^{2} - 2$ $)$

Hence, this is the answer.

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