Question

pq4-p(p-q)4 factorise it

Answer 1

Please see the attachment

Answer 2

p q^4 - p (p-q)^4 = p^2 (2q - p) (2q^2 + p ^2 - 2pq)

Explanation:

p q^4 - p (p-q)^4

Taking p common, we get; p [ q^4 - (p-q)^4 ]

= p [ (q^2)^2 - {(p-q)^2}^2 ]  ; now it is in the format a^2-b^2

= p [ {q^2 + (p-q)^2} {q^2 - (p-q)^2} ]      ; as we know:  a^2-b^2 = (a + b)(a - b)

= p [ {q^2 + (p-q)^2} { (q + (p-q) ) ( q - (p-q)}  ] , As a^2-b^2 = (a + b)(a - b)

= p [ {q^2 + (p-q)^2} { (q + p - q) ( q - p +q)}  ]

= p [ {q^2 + (p-q)^2} {  p (2q - p)}  ]

= p^2 [ {q^2 + p ^2 + q ^2 - 2pq} (2q - p)

= p^2 (2q - p) (2q^2 + p ^2 - 2pq) (Answer)