Question

X^4 - ( X - Z ) ^4
factorise using identities ..
plz answer I will mark it the branlist​

Answer 1

x^4-(x-z)^4 is the given expression

let (x-z) = y

So, x^4 - y^4 = (x^2)^2 - (y^2)^2 = {(x^2) + (y^2)}

{(x^2) - (y^2)} = {(x^2) + (y^2)} (x+y) (x-y)

Put the value of y into the found expression

= {(x^2) + (y^2)} (x+y) (x-y)

= {x^2+(x-z)^2} (x+x-z) (x-x+z)

= (x^2+x^2+z^2-2xz) (2x-z) (z)

= (2x^2+z^2-2xz) (2x-z)(z)

THE FINAL ANSWER IS

$\boxed {(2x^2+z^2-2xz) (2x-z)(z)}$

Answer 2

Question :

Factorize

$x {}^{4} - (x - z) {}^{4}$

Formula used :

${\purple{\boxed{\large{\bold{x {}^{2} - y {}^{2} = ( x + y)(x - y)}}}}}$

Solution :

$x {}^{4} - (x - z) {}^{4}$

$=( x {}^{2} ) {}^{2} -( (x - z) {}^{2} ) {}^{2}$

[ use formula :x²- y² = (x+y ) (x-y) ]

$= (x {}^{2} + (x - z) {}^{2} ) + (x {}^{2} - (x - z) {}^{2} )$

$=( x {}^{2} + (x - z) {}^{2} )(x + (x - z))(x - (x - z))$

$= (x {}^{2} + x {}^{2} + z {}^{2} - 2xz)(2x - z)(z)$

$= (2x { }^{2} + z {}^{2} - 2xz)(2x - z)(z)$

which is the required solution!