If p+q+r=0 then show that (p+r)³= 3pqr
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p+q+r=0
= p+q= -r
= (p+q)³= (-r)³
= p³+q³+3p²q+3pq²= -r³
= p³+q³+r³ = -3pq(p+q)
= p³+q³+r³=3pqr
= p+q= -r
= (p+q)³= (-r)³
= p³+q³+3p²q+3pq²= -r³
= p³+q³+r³ = -3pq(p+q)
= p³+q³+r³=3pqr
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