(p+1)^3 evaluate with identities
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( p + 1 )³
Using Identity
( a + b )³ = a³ + b³ + 3ab( a + b )
→ a³ + b³ + 3a²b + 3ab²
( p + 1 )³
→ ( p )³ + ( 1 )³ + 3 (p) (1) ( p + 1 )
→ p³ + 1 + 3p( p + 1 )
→ p³ + 1 + 3p² + 3p
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