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a³-b³+1+3ab......................Factorise it please.
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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
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
Yeah.
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Can't you use any algebraic identity?
hope this helps
Which standard are you?
there is no direct algebraic identity to factorise that
12th
Great........
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0
Answer:
(a-b) (a2+b2+ab)+1+3ab
Step-by-step explanation:
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