展開する ((2x+3))/((x^2+x+1)^2(x-1)^3(x+1)^2)

解

展開する

\frac{( 2 x + 3 )}{( x ^{2} + x + 1 ) ^{2} ( x - 1 ) ^{3} ( x + 1 ) ^{2}}

\frac{2 x + 3}{x ^{9} + x ^{8} - x ^{7} - 3 x ^{6} - 2 x ^{5} + 2 x ^{4} + 3 x ^{3} + x ^{2} - x - 1}

解答ステップ

\frac{2 x + 3}{( x ^{2} + x + 1 ) ^{2} ( x - 1 ) ^{3} ( x + 1 ) ^{2}}

(x^{2} + x + 1)^{2} = x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 1

= \frac{2 x + 3}{( x ^{4} + 2 x ^{3} + 3 x ^{2} + 2 x + 1 ) ( x - 1 ) ^{3} ( x + 1 ) ^{2}}

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

= \frac{2 x + 3}{( x ^{4} + 2 x ^{3} + 3 x ^{2} + 2 x + 1 ) ( x ^{3} - 3 x ^{2} + 3 x - 1 ) ( x + 1 ) ^{2}}

(x + 1)^{2} = x^{2} + 2 x + 1

= \frac{2 x + 3}{( x ^{4} + 2 x ^{3} + 3 x ^{2} + 2 x + 1 ) ( x ^{3} - 3 x ^{2} + 3 x - 1 ) ( x ^{2} + 2 x + 1 )}

拡張

(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 1) (x^{3} - 3 x^{2} + 3 x - 1) (x^{2} + 2 x + 1) : x^{9} + x^{8} - x^{7} - 3 x^{6} - 2 x^{5} + 2 x^{4} + 3 x^{3} + x^{2} - x - 1

= \frac{2 x + 3}{x ^{9} + x ^{8} - x ^{7} - 3 x ^{6} - 2 x ^{5} + 2 x ^{4} + 3 x ^{3} + x ^{2} - x - 1}