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

Question

Saksham Tolani · Accepted Answer

Dear Student,
Assuming equation to be x4 – (x
– z)4
The expression x4 – (x –
z)4 can be factorised as,
x4 – (x –
z)4
= (x2)2 – {(x
– z)2}2
= [x2 – (x –
z)2] [x2 + (x –
z)2] [a2 –
b2 = (a – b) (a +
b)]
= [x – (x – z)] [x +
(x – z)] [x2 +
x2 + z2 –
2xz]
[a2 – b2 = (a
– b) (a + b), (a –
b)2 = a2 + b2
–2ab]
= [x – x + z)] [2x –
z] [x2 + x2 +
z2 – 2xz]
= z (2x – z) [2x2
+ z2 – 2xz]

Regards!

factorise; x4-[x-z]4 with help of suitable identities

_{^{x4−(x−z)4

=(x2)2−{(x−z)2}2

=(x2+(x−z)2)(x2−(x−z)2)2

=(x2+x2−2xz+z2)(x+x−z)(x−x+z)

=(2x2−2xz+z2)(2x−z)(z)}}

x4−(x−z)4 =(x2)2−{(x−z)2}2 =(x2+(x−z)2)(x2−(x−z)2)2 =(x2+x2−2xz+z2)(x+x−z)(x−x+z) =(2x2−2xz+z2)(2x−z)(z)

factorise; x4-[x-z]4 with help of suitable identities

_{^{x4−(x−z)4

=(x2)2−{(x−z)2}2

=(x2+(x−z)2)(x2−(x−z)2)2

=(x2+x2−2xz+z2)(x+x−z)(x−x+z)

=(2x2−2xz+z2)(2x−z)(z)}}