A) \[(ac+bd+bc+ad)(ac-bd-bc-ad)\]
B) \[(ac-bd-bc+ad)(ac+bd+dc+ab)\]
C) \[(ac-bd+bc+ad)(ac-bd=-bc-ad)\]
D) \[(ac-bd-bc-ad)(ac+bd+dc+ab)\]
Correct Answer: C
Solution :
\[({{a}^{2}}-{{b}^{2}})({{c}^{2}}-{{d}^{2}})-4abcd\] \[={{a}^{2}}{{c}^{2}}-{{a}^{2}}{{d}^{2}}-{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}-4abcd\] \[=({{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}-2abcd)-({{b}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+2abcd)\] \[={{(ac-bd)}^{2}}-{{(bc+ad)}^{2}}\] \[=(ac-bd+bc+ad)(ac-bd-bc-ad)\]You need to login to perform this action.
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