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Question

Factorize the equation $$\left ( a^{2}-1 \right )\left ( b^{2} -1\right )+4ab$$


A
(a21)(b21)+4ab=(ab+1+ab)(abb)
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B
(a21)(b21)+4ab=(ab+1ab)(ab+1a+b)
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C
(a21)(b21)+4ab=(ab+1+ab)(ab+1)
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D
(a31)(b21)+4ab=(ab+1+ab)(ab+1ab)
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Solution

The correct option is B $$\left ( a^{2}-1 \right )\left ( b^{2}-1 \right )+4ab= \left ( ab+1-a-b \right )\left ( ab+1-a+b \right )$$
    $$\left ( a^{2}-1 \right )\left ( b^{2} -1\right )+4ab$$
$$=a^2b^2-a^2-b^2+1+4ab$$$$=a^2b^2+2ab+1-(a^2+b^2-2ab)$$
$$=(ab+1)^2-(a-b)^2$$
$$=(ab+1+a-b)(ab+1-a+b)$$

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