(4 – p)3 =(4)3 – [ 3 ×(4)2 × p ] + [ 3 ×(4)×(p)2 ] –(p)3{ ∵(a – b)3 = a3 – 3a2b + 3ab2 – b3 }= 64 – [3 × 6 × p ] + [ 3 × 4 × p2 ] – p3= 64 – 48p + 12p2 – p3

Q. 1 B4.1( 14 Votes )

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Class 8th MHB - Mathematics (New)5. Expansion Formulae

Answer :

(4 – p)³ =(4)³ – [ 3 ×(4)² × p ] + [ 3 ×(4)×(p)² ] –(p)³

{ ∵(a – b)³ = a³ – 3a²b + 3ab² – b³ }

= 64 – [3 × 6 × p ] + [ 3 × 4 × p² ] – p³

= 64 – 48p + 12p² – p³

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