Bagaimana cara menentukan perpangkatan bentuk aljabar berikut? Jelaskan.
a. (a + b)5
b. (a + b + c)2,
c. (a + b – c)2,
d. (a – b + c)2,
e. (a – b – c)2
Cara Penyelesaian
a. (a + b)⁵ => menggunakan Segitiga Pascal
= 1(a)⁵ (b)⁰ + 5 (a)⁴ (b)¹ + 10 (a)³ (b)² + 10 (a)² (b)³ + 5 (a)¹ (b)⁴ + 1(a)⁰ (b)⁵
= a⁵ + 5 a⁴ b + 10 a³ b² + 10 a² b³ + 5 a b⁴ + b⁵
b. (a + b + c)² = (a + b + c) (a + b + c)
= a² + ab + ac + ab + b² + bc + ac + bc + c²
= a² + b² + c² + 2ab + 2ac + 2bc
= a² + b² + c² + 2 (ab + ac + bc)
c. (a + b – c)² = (a + b – c) (a + b – c)
= a² + ab – ac + ab + b² – bc – ac – bc + c²
= a² + b² + c² + 2ab – 2ac – 2bc
= a² + b² + c² + 2 (ab – ac – bc)
d. (a – b + c)² = (a – b + c) (a – b + c)
= a² – ab + ac – ab + b² – bc + ac – bc + c²
= a² + b² + c² – 2ab + 2ac – 2bc
= a² + b² + c² – 2 (ab – ac + bc)
e. (a – b – c)² = (a – b – c) (a – b – c)
= a² – ab – ac – ab + b² + bc – ac + bc + c²
= a² + b² + c² – 2ab – 2ac + 2bc
= a² + b² + c² – 2 (ab + ac – bc)
= a² + b² + c² – 2 (ab + ac – bc)
