因数 a^5b^{10}-c^{15}d^5

解

因数

a^{5} b^{10} - c^{15} d^{5}

(a b^{2} - c^{3} d) (a^{4} b^{8} + a^{3} b^{6} c^{3} d + a^{2} b^{4} c^{6} d^{2} + a b^{2} c^{9} d^{3} + c^{12} d^{4})

解答ステップ

a^{5} b^{10} - c^{15} d^{5}

a^{5} b^{10}

を書き換え

(a b^{2})^{5}

c^{15} d^{5}

を書き換え

(c^{3} d)^{5}

= (a b^{2})^{5} - (c^{3} d)^{5}

因数分解の規則を適用する:

x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + \dots + x y^{n - 2} y^{n - 1})

(a b^{2})^{5} - (c^{3} d)^{5} = (a b^{2} - c^{3} d) ((a b^{2})^{4} + c^{3} d (a b^{2})^{3} + (c^{3} d)^{2} (a b^{2})^{2} + (c^{3} d)^{3} a b^{2} + (c^{3} d)^{4})

= (a b^{2} - c^{3} d) ((a b^{2})^{4} + c^{3} d (a b^{2})^{3} + (c^{3} d)^{2} (a b^{2})^{2} + (c^{3} d)^{3} a b^{2} + (c^{3} d)^{4})

簡素化

(a b^{2} - c^{3} d) ((a b^{2})^{4} + c^{3} d (a b^{2})^{3} + (c^{3} d)^{2} (a b^{2})^{2} + (c^{3} d)^{3} a b^{2} + (c^{3} d)^{4}) : (a b^{2} - c^{3} d) (a^{4} b^{8} + a^{3} b^{6} c^{3} d + a^{2} b^{4} c^{6} d^{2} + a b^{2} c^{9} d^{3} + c^{12} d^{4})

= (a b^{2} - c^{3} d) (a^{4} b^{8} + a^{3} b^{6} c^{3} d + a^{2} b^{4} c^{6} d^{2} + a b^{2} c^{9} d^{3} + c^{12} d^{4})