Question

can anyone plz ...factorise the given problem

Answer 1

Start by seeing that the first two terms are the difference of two cubes 
a³ – b³ ≡ (a – b)(a² + ab + b²) 

Hence 
a³ – b³ + 1 + 3ab = (a – b)(a² + ab + b²) + 1 + 3ab 

Now add 1 inside the first brackets and subtract 1 times the second brackets to leave the expression unchanged in value 
a³ – b³ + 1 + 3ab = (a – b + 1)(a² + ab + b²) –1(a² + ab + b²) + 1 + 3ab 
a³ – b³ + 1 + 3ab = (a – b + 1)(a² + ab + b²) – a² – ab – b² + 1 + 3ab 
a³ – b³ + 1 + 3ab = (a – b + 1)(a² + ab + b²) – (a² + ab + b² – 1 – 3ab) 
a³ – b³ + 1 + 3ab = (a – b + 1)(a² + ab + b²) – (a² + b² – 1 – 2ab) 

Recall that (a – b)² ≡ a² – 2ab + b² 
a³ – b³ + 1 + 3ab = (a – b + 1)(a² + ab + b²) – ((a – b)² – 1) 

Now see the difference of two squares 
((a – b)² – 1) = ((a – b)² – 1²) = (a – b + 1)(a – b – 1) 

Hence 
a³ – b³ + 1 + 3ab = (a – b + 1)(a² + ab + b²) – (a – b + 1)(a – b – 1) 

Take out the common factor which is (a – b + 1) 
a³ – b³ + 1 + 3ab = (a – b + 1)[(a² + ab + b²) – (a – b – 1)] 

Simplify 
a³ – b³ + 1 + 3ab = (a – b + 1)(a² + ab + b² – a + b + 1)

Answer 2

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