Question

factorise( x-1)^3+(x-2)^3+(3-2x)^3
pls help me guyzzz ....its urgent to answer​

Answer 1

Answer:

let a=x-1

let b = x-2

let c=3-2x

if a +b+ c=0...then a³+b³+c³=3abc

X-1+x-2+3-2x

=0

therefore (x-1)³ +(x-2)³+(3-2x)³=3(x-1)(x-2)(3-2x)

Answer 2

Answer:

3(x-1)(x-2)(3-2x)

Step-by-step explanation:

Let (x-1)be 'a'

(x-2) be 'b'

(3-2x) be'c'

(x-1)^3+(x-2)^3+(3-2x)^3=a^3+b^3+c^3

Now ,

a+b+c=0 [Here , (x-1)+(x-2)+(3-2x)=0

a+b=-c x-1+x-2+3-2x=0

2x-3+3-2x=0

0=0]

Cubing both sides

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

a^3+b^3+3ab(a+b)=-c^3

a^3+b^3+3ab(-c)=-c^3

a^3+b^3+c^3-3abc=0

a^3+b^3+c^3=3abc

Putting values of a, b, c

(x-1)^3+(x-2)^3+(3-2x)=3(x-1)(x-2)(3-2x)