# Pleeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeease help! a³-b³+1+3ab......................Factorise it please.

1
by Deleted account
are u of 9th?
yes
and u
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ohk sorry bro

2015-07-12T11:42:38+05:30

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A³ - b³ + 1 + 3ab
a³ – b³ + 1 + 3ab = (a – b)(a² + ab + b²) + 1 + 3ab

adding 1 inside the first brackets and subtract 1 times the second brackets to leave the expression unchanged in value

= (a – b + 1)(a² + ab + b²) –1(a² + ab + b²) + 1 + 3ab
= (a – b + 1)(a² + ab + b²) – a² – ab – b² + 1 + 3ab
= (a – b + 1)(a² + ab + b²) – (a² + ab + b² – 1 – 3ab)
= (a – b + 1)(a² + ab + b²) – (a² + b² – 1 – 2ab)

as (a – b)² ≡ a² – 2ab + b²

= (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
= (a – b + 1)(a² + ab + b²) – (a – b + 1)(a – b – 1)

= (a – b + 1)[(a² + ab + b²) – (a – b – 1)]

= (a – b + 1)(a² + ab + b² – a + b + 1)ANSWER

hope this helps
Which standard are you?
there is no direct algebraic identity to factorise that
12th
Great........