Question

Factorise :
1 - ab - abc + a2b2c-a + a2b + a2bc-a3b2c
Options​

Answer 1

repeat the question

(a+b+c)2=a2+b2+c2+ab+bc+ca

Answer 2

Given:

Given expression is:

$1 - ab - abc + a^2b^2c-a + a^2b + a^2bc-a^3b^2c$

To find:

We need to factorize the given expression.

Solution:

Consider the given expression,

$\Rightarrow 1 - ab - abc + a^2b^2c-a + a^2b + a^2bc-a^3b^2c$

Rearranging the terms, we get

$\Rightarrow 1 - ab - abc -a + a^2b + a^2b^2c+ a^2bc-a^3b^2c$

Taking $a^2bc$ as common in the last three terms,

$\Rightarrow 1 - ab - abc -a + a^2b + a^2bc(b+ 1-ab)$

Rearranging the terms, we get

$\Rightarrow 1 - abc - ab -a + a^2b + a^2bc(b+ 1-ab)$

Taking $-a$ common in some of the terms,

$\Rightarrow 1 - abc - a(b +1 - ab) + a^2bc(b+ 1-ab)$

Taking $b +1 - ab$ as common in the above expression,

$\Rightarrow 1 - abc+(b +1 - ab) (-a+ a^2bc)$

Taking $-a$ common in some of the terms,

$\Rightarrow 1 - abc+(b +1 - ab) (-a(1-abc))$

Simplifying the terms, we get

$\Rightarrow 1 - abc -a(b +1 - ab) (1-abc)$

Taking $1-abc$ as common in the above expression,

$\Rightarrow (1 - abc)(1 -a(b +1 - ab))$

The above expression cannot be simplified further.

Therefore, the required factorized form is

$(1 - abc)(1 -a(b +1 - ab))$.