展開する (x^3+b+1)^3*x^2

解

展開する

(x^{3} + b + 1)^{3} \cdot x^{2}

x^{11} + 3 b x^{8} + 3 x^{8} + 3 b^{2} x^{5} + 6 b x^{5} + 3 x^{5} + b^{3} x^{2} + 3 b^{2} x^{2} + 3 b x^{2} + x^{2}

解答ステップ

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

(x^{3} + b + 1)^{3} = x^{9} + 3 b x^{6} + 3 x^{6} + 3 b^{2} x^{3} + 6 b x^{3} + 3 x^{3} + b^{3} + 3 b^{2} + 3 b + 1

= x^{2} (x^{9} + 3 b x^{6} + 3 x^{6} + 3 b^{2} x^{3} + 6 b x^{3} + 3 x^{3} + b^{3} + 3 b^{2} + 3 b + 1)

= x^{2} (x^{9} + 3 b x^{6} + 3 x^{6} + 3 b^{2} x^{3} + 6 b x^{3} + 3 x^{3} + b^{3} + 3 b^{2} + 3 b + 1)

括弧を分配する

= x^{2} x^{9} + x^{2} \cdot 3 b x^{6} + x^{2} \cdot 3 x^{6} + x^{2} \cdot 3 b^{2} x^{3} + x^{2} \cdot 6 b x^{3} + x^{2} \cdot 3 x^{3} + x^{2} b^{3} + x^{2} \cdot 3 b^{2} + x^{2} \cdot 3 b + x^{2} \cdot 1

= x^{9} x^{2} + 3 b x^{6} x^{2} + 3 x^{6} x^{2} + 3 b^{2} x^{3} x^{2} + 6 b x^{3} x^{2} + 3 x^{3} x^{2} + b^{3} x^{2} + 3 b^{2} x^{2} + 3 b x^{2} + 1 \cdot x^{2}

簡素化

x^{9} x^{2} + 3 b x^{6} x^{2} + 3 x^{6} x^{2} + 3 b^{2} x^{3} x^{2} + 6 b x^{3} x^{2} + 3 x^{3} x^{2} + b^{3} x^{2} + 3 b^{2} x^{2} + 3 b x^{2} + 1 \cdot x^{2} : x^{11} + 3 b x^{8} + 3 x^{8} + 3 b^{2} x^{5} + 6 b x^{5} + 3 x^{5} + b^{3} x^{2} + 3 b^{2} x^{2} + 3 b x^{2} + x^{2}

= x^{11} + 3 b x^{8} + 3 x^{8} + 3 b^{2} x^{5} + 6 b x^{5} + 3 x^{5} + b^{3} x^{2} + 3 b^{2} x^{2} + 3 b x^{2} + x^{2}