HELP OR DIE.a factorisation question?

a^2 - b^2 ≡ (a + b)(a - b)

x^3 - 121x
= x(x^2 - 121)
= x(x^2 - 11^2)
= x(x + 11)(x - 11)

let f(x)=x^3-121x

find f(0)
You'll get 0
So x=0
x is a factor

0
| 1 0 -121 0
| 0 0 0
---------------------
1 0 -121 | 0

so x^3-121x can be written as
x^3-121x = (x)(x^2-121)

x^2-121= x^2 -11^2
=(x+11)(x-11)

Therefore finally we can write

x^3-121x = (x)(x+11)(x-11)

Thats it

A rather strange presentation !

However :-

x ³ - 121 x = x ( x ² - 121 ) = x ( x - 11 ) ( x + 11 )

step 1:-
taking "x" common from the expression than it will become

= x(x^2 - 121)

making pairs in the form of a^2 - b^2

= x ( (x)^2 - (11)^2)

as we know that a^2 - b^2 = (a+b)(a-b)

then this expression will be written in the follwing form

=x(x-11)(x+11)