展開する (x+1)^3(6-x)^2\sqrt[3]{2x+1}

解

展開する

(x + 1)^{3} (6 - x)^{2} 3 2 x + 1

x^{5} 3 2 x + 1 - 9 x^{4} 3 2 x + 1 + 3 x^{3} 3 2 x + 1 + 73 x^{2} 3 2 x + 1 + 96 x 3 2 x + 1 + 36 3 2 x + 1

解答ステップ

(x + 1)^{3} (6 - x)^{2} 3 2 x + 1

拡張

(6 - x)^{2} : 36 - 12 x + x^{2}

= (x + 1)^{3} (36 - 12 x + x^{2}) 3 2 x + 1

= (x + 1)^{3} 3 2 x + 1 (36 - 12 x + x^{2})

拡張

(x + 1)^{3} : x^{3} + 3 x^{2} + 3 x + 1

= (x^{3} + 3 x^{2} + 3 x + 1) 3 2 x + 1 (36 - 12 x + x^{2})

= 3 2 x + 1 (x^{3} + 3 x^{2} + 3 x + 1) (36 - 12 x + x^{2})

拡張

(x^{3} + 3 x^{2} + 3 x + 1) (36 - 12 x + x^{2}) : x^{5} - 9 x^{4} + 3 x^{3} + 73 x^{2} + 96 x + 36

= 3 2 x + 1 (x^{5} - 9 x^{4} + 3 x^{3} + 73 x^{2} + 96 x + 36)

括弧を分配する

= 3 2 x + 1 x^{5} + 3 2 x + 1 (- 9 x^{4}) + 3 2 x + 1 \cdot 3 x^{3} + 3 2 x + 1 \cdot 73 x^{2} + 3 2 x + 1 \cdot 96 x + 3 2 x + 1 \cdot 36

簡素化

3 2 x + 1 x^{5} + 3 2 x + 1 (- 9 x^{4}) + 3 2 x + 1 \cdot 3 x^{3} + 3 2 x + 1 \cdot 73 x^{2} + 3 2 x + 1 \cdot 96 x + 3 2 x + 1 \cdot 36 : x^{5} 3 2 x + 1 - 9 x^{4} 3 2 x + 1 + 3 x^{3} 3 2 x + 1 + 73 x^{2} 3 2 x + 1 + 96 x 3 2 x + 1 + 36 3 2 x + 1

= x^{5} 3 2 x + 1 - 9 x^{4} 3 2 x + 1 + 3 x^{3} 3 2 x + 1 + 73 x^{2} 3 2 x + 1 + 96 x 3 2 x + 1 + 36 3 2 x + 1