Question

3a2 - b2 - c2 + 2ab - 2bc + 2ca -- Factorise

Answer 1

Please see the attachment

Answer 2

GIVEN :

The expression is $3a^2-b^2-c^2+2ab-2bc+2ca$

TO FIND :

The factors of the given expression by using factorise

SOLUTION :

Given expression is $3a^2-b^2-c^2+2ab-2bc+2ca$

Now factorise the given expression as below

$3a^2-b^2-c^2+2ab-2bc+2ca$

Adding and subtracting $a^2$ in the above expression we get

$=3a^2+a^2-a^2-b^2-c^2+2ab-2bc+2ca$

Adding the like terms we get,

$=4a^2-a^2-b^2-c^2+2ab-2bc+2ca$

$=4a^2-[a^2+b^2+c^2-2ab+2bc-2ca]$

$=4a^2-[a^2-b^2-c^2+2(a)(-b)+2(-b)(-c)+2(-c)(a)]$

By using the algebraic formula :

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

$=4a^2-(a-b-c)^2$  (since here a=a , b=-b and c=-c )

$=(2a)^2-(a-b-c)^2$

By using the algebraic formula :

$(a+b)(a-b)=a^2-b^2$

$=(2a+a-b-c)(2a-(a-b-c))$

$=(2a+a-b-c)(2a-a+b+c)$

$=(3a-b-c)(a+b+c)$

∴ $3a^2-b^2-c^2+2ab-2bc+2ca=(3a-b-c)(a+b+c)$

∴ the given expression $3a^2-b^2-c^2+2ab-2bc+2ca$ is factored as (3a-b-c)(a+b+c)