factorise by using identity
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Hi !!
( a + b )² - 49 ( a - b )²
Using identity , ( a + b)² = a² + b² + 2ab and ( a - b)² = a² + b² - 2ab.
Therefore,
( a)² + ( b)² + 2ab - 49 [ ( a² + b² - 2ab ) ]
a² + b² + 2ab - 49a² - 49b² + 98ab
a² - 49a² + b² - 49b² + 2ab + 98ab
-48a² - 48b² + 100ab
-4 ( 12a² + 12b² 25ab) [ Answer ]
( a + b )² - 49 ( a - b )²
Using identity , ( a + b)² = a² + b² + 2ab and ( a - b)² = a² + b² - 2ab.
Therefore,
( a)² + ( b)² + 2ab - 49 [ ( a² + b² - 2ab ) ]
a² + b² + 2ab - 49a² - 49b² + 98ab
a² - 49a² + b² - 49b² + 2ab + 98ab
-48a² - 48b² + 100ab
-4 ( 12a² + 12b² 25ab) [ Answer ]
mayank1797:
wrong answer please read the statement carefully
Answered by
0
Answer will be 4(-12a²-12b²+20ab).
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