Question

FACTORISE:-
A³-B³+1+3AB

Answer 1

Answer:

(a – b + 1)(a² + ab + b² – a + b + 1)

Step-by-step explanation:

Hey there,

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)

Hope this helps! 

Answer 2

Answer:

using identity ->

x^3+y^3+z^3–3.x.y.z =(x+y+z) (x^2+y^2+z^2-xy-yz-zx).

a³-b³+1+3ab

=(a)^3 +(-b)^3 + (1)^3 -3.a.(-b).1

=[a+(-b,)+1] [(a)^2+((- b)^2+(1)^2-a.(- b)-(-b).1- 1.a]

=(a-b+1) (a^2+b^2+1+a.b+b-a)