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प्रश्न
Factorise the following:
(1 – x2) (1 – y2) + 4xy
योग
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उत्तर
Given: (1 – x2) (1 – y2) + 4xy
Step-wise calculation:
Step 1: Expand (1 – x2)(1 – y2)
(1 – x2)(1 – y2)
= 1 – y2 – x2 + x2y2
So the expression becomes:
1 – x2 – y2 + x2y2 + 4xy
Rearrange:
x2y2 – x2 – y2 + 1 + 4xy
Step 2: Recognize a useful identity
Notice that:
x2y2 – x2 – y2 + 1
= (xy – 1)2 – (x – y)2
Check:
(xy – 1)2 = x2y2 – 2xy + 1
(x – y)2 = x2 – 2xy + y2
Subtract:
(xy – 1)2 – (x – y)2
= x2y2 – x2 – y2 + 1
So the original expression becomes:
(xy – 1)2 – (x – y)2 + 4xy
But notice:
4xy = 4xy – (–2xy + 2xy) = 4xy
We group terms differently:
x2y2 – x2 – y2 + 1 + 4xy
= (xy + x – y + 1)(xy – x + y + 1)
Final Answer
(1 + xy + x – y)(1 + xy – x + y)
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