Question

factorise: a⁴b⁴+1+a²b²-c²+2abc​

Answer 1

Answer:

The given question is solved in the uploaded photo

Answer 2

Answer:

( a²b² + 1 + ab - c ) * ( a²b² + 1 - ab + c )

Step-by-step explanation:

a⁴b⁴ + 1 + a²b² - c² + 2abc

a⁴b⁴ + 2a²b² + 1² - a²b² - c² + 2abc -> ( adding and subtracting a²b² )

If you see carefully you will be seeing two identity being applied :

( a + b )² = a² + b² + 2ab ------------- in ( a⁴b⁴ + 2a²b² + 1² )

and

( a - b )² = a² + b² - 2ab -------------- in ( - a²b² - c² + 2abc ) = ( a²b² + c² - 2abc )

Thus now you can solve it... Further you will be Solving it with one more identity... just wait

the above expression will be:

( a²b² + 1 )² - ( ab - c )²

Now which identity is being applied here?

yes, it is ---- ( a² - b² ) = ( a + b ) ( a - b )

therefore,

( a²b² + 1 + ( ab - c ) ) * ( a²b² + 1 - ( ab - c ) )

= ( a²b² + 1 + ab - c ) * ( a²b² + 1 - ab + c )