factorise=(p-q)^3 + (q-r)^3 + (r-p)^3
Answers
Answered by
6
here,
( p - q ) + (q - r ) + ( r - p ) =0
as we know, that,
if a + b + c = 0
then,
a^3 + b^3 + c^3 = 3abc
therefore,
( p - q )^3+(q - r )^3 + (r - p)^3 = 3(p-q)(q-r)(r-p)
( p - q ) + (q - r ) + ( r - p ) =0
as we know, that,
if a + b + c = 0
then,
a^3 + b^3 + c^3 = 3abc
therefore,
( p - q )^3+(q - r )^3 + (r - p)^3 = 3(p-q)(q-r)(r-p)
Answered by
12
(p-q)³+(q-r)³+(r-p)³
=p³-q³-3p²q+3pq²+q³-r³-3q²r+3qr²+r³-p³-3r²p+3rp²
=3pq²+3qr²+3rp²-(3qp²+3rq²+3pr²)
=3(p-q)(q-r)(r-p)
=p³-q³-3p²q+3pq²+q³-r³-3q²r+3qr²+r³-p³-3r²p+3rp²
=3pq²+3qr²+3rp²-(3qp²+3rq²+3pr²)
=3(p-q)(q-r)(r-p)
Similar questions