factorise: (p-q)^3+(q-r)^3+(r-p)^3
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This question can be solved by the identity ;
'When a+ b+ c = 0 , a³ + b³ + c³ = 3abc '
Here, a = p-q , b = q-r , c= r-p
And, a+b+c= p-q +q-r +r-p = 0.
So, (p-q)³ + (q-r)³ + (r-p)³ = 3(p-q)(q-r)(r-p)
'When a+ b+ c = 0 , a³ + b³ + c³ = 3abc '
Here, a = p-q , b = q-r , c= r-p
And, a+b+c= p-q +q-r +r-p = 0.
So, (p-q)³ + (q-r)³ + (r-p)³ = 3(p-q)(q-r)(r-p)
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