# factorise; x*4-[x-z]*4 with help of suitable identities

Dear Student,

Assuming equation to be *x*^{4} – (*x* – *z*)^{4}

The expression

*x*^{4}– (*x*–*z*)^{4}can be factorised as,*x*

^{4}– (

*x*–

*z*)

^{4}

= (

*x*^{2})^{2}– {(*x*–*z*)^{2}}^{2}= [

*x*^{2}– (*x*–*z*)^{2}] [*x*^{2}+ (*x*–*z*)^{2}] [*a*^{2}–*b*^{2}= (*a*–*b*) (*a*+*b*)]= [

*x*– (*x*–*z*)] [*x*+ (*x*–*z*)] [*x*^{2}+*x*^{2}+*z*^{2}– 2*xz*] [

*a*^{2}–*b*^{2}= (*a*–*b*) (*a*+*b*), (*a*–*b*)^{2}=*a*^{2}+*b*^{2}–2*ab*]= [

*x*–*x*+*z*)] [2*x*–*z*] [*x*^{2}+*x*^{2}+*z*^{2}– 2*xz*]=

*z*(2*x*–*z*) [2*x*^{2 }+*z*^{2}– 2*xz*]Regards!

**
**