Question

factorise  (p-q)³+(q-r)³+(r-p)³​

Answer 1

Answer:

p³-q³+q³-r³+r³-p³

= 0

This may be the answer.

Answer 2

(p-q)^3+ ( q-r)^3+(r-p)^3

= p^3-3pq(p-q)-q^3+ q^3-3qr(q-r)-r^3+ r^3-3pr(r-p)-p^3

= -3p^2q+3q^2p-3q^2r+3qr^2-3pr^2+3p^2r

=- 3 [p^2q- qr^2- pq^2+ q^2r-p^2r+pr^2]

= -3 [ q( p^2 - r^2) -q^2(p- r)-pr(p- r)]

= - 3 [ q( p-r) ( p+r)- q^2 (p-r)- pr (p-r)]

= -3( p- r) [ q(p+r) - q^2- pr)

= -3 ( p-r )[ qp + qr-q^2 -pr]

= - 3( p-r ) [ qp -q^2 -pr + qr]

= - 3 (p- r) ( q( p- q) - r( p- q)]

= - 3 ( p- r) (p- q) ( q- r)

= 3( p- q) ( q- r) ( r- p)

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perhaps it is correct ..