因数 64j^6-k^9

解

因数

64 j^{6} - k^{9}

(4 j^{2} - k^{3}) (16 j^{4} + 4 j^{2} k^{3} + k^{6})

解答ステップ

64 j^{6} - k^{9}

指数の規則を適用する

= (4 j^{2})^{3} - (k^{3})^{3}

立方数の差の公式を適用する:

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

(4 j^{2})^{3} - (k^{3})^{3} = (4 j^{2} - k^{3}) ((4 j^{2})^{2} + 4 j^{2} k^{3} + (k^{3})^{2})

= (4 j^{2} - k^{3}) ((4 j^{2})^{2} + 4 j^{2} k^{3} + (k^{3})^{2})

簡素化

(4 j^{2})^{2} + 4 j^{2} k^{3} + (k^{3})^{2} : 16 j^{4} + 4 j^{2} k^{3} + k^{6}

= (4 j^{2} - k^{3}) (16 j^{4} + 4 j^{2} k^{3} + k^{6})