Jabarkanlah perpangkatan bentuk aljabar brikut (pilih cara perkalian berulang atau dngn segitiga pascal) a.(4 m-3 n)(pangkat4) b.(4 m+3 n)(pankat4)

a. (4m - 3n)^4
= 1×(4m)^4×(-3n)^0 + 4×(4m)^3×(-3n)^1 + 6×(4m)^2×(-3n)^2 + 4×(4m)^1×(-3n)^3 + 1×(4m)^0×(-3n)^4

= (1×4^4m^4 × 1) + (4×4^3m^3×(-3)^1n^1) + (6×4^2m^2×(-3)^2n^2) + (4×4^1m^1×(-3)^3n^3) + (1×1×(-3)^4n^4)

= 256m^4 + (4×64×(-3) m^3n) + (6×16×9 m^2n^2) + (4×1×(-27) mn^3) + 81 n^4

= 256m^4 + (-768m^3n) + 864m^2n^2 + (- 108mn^3) + 81 n^4

= 256 m^4 - 768 m^3n + 864 m^2n^2 - 108 mn^3 + 81 n^4

---------☆☆☆------☆☆☆---------☆☆☆--------

b. (4m + 3n) ^4
= 1×(4m)^4×(3n)^0 + 4×(4m)^3×(3n)^1 + 6×(4m)^2×(3n)^2 + 4×(4m)^1×(3n)^3 + 1×(4m)^0×(3n)^4

= (1×4^4m^4 × 1) + (4×4^3m^3×3^1n^1) + (6×4^2m^2×3^2n^2) + (4×4^1m^1×3^3n^3) + (1×1×3^4n^4)

= 256m^4 + (4×64×3m^3n) + (6×16×9 m^2n^2) + (4×1×27 mn^3) + 81 n^4

= 256m^4 + 768m^3n + 864m^2n^2 + 108mn^3 + 81 n^4


......semoga membantu.....