What is expand (-2t^5-2t)/((t^2+1)^2(t^2-1)^3) ?

The solution to expand (-2t^5-2t)/((t^2+1)^2(t^2-1)^3) is (-2t^5-2t)/(t^{10)-t^8-2t^6+2t^4+t^2-1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (-2t^5-2t)/((t^2+1)^2(t^2-1)^3)

expand

\frac{- 2 t ^{5} - 2 t}{( t ^{2} + 1 ) ^{2} ( t ^{2} - 1 ) ^{3}}

\frac{- 2 t ^{5} - 2 t}{t ^{10} - t ^{8} - 2 t ^{6} + 2 t ^{4} + t ^{2} - 1}

Solution steps

\frac{- 2 t ^{5} - 2 t}{( t ^{2} + 1 ) ^{2} ( t ^{2} - 1 ) ^{3}}

(t^{2} + 1)^{2} = t^{4} + 2 t^{2} + 1

= \frac{- 2 t ^{5} - 2 t}{( t ^{4} + 2 t ^{2} + 1 ) ( t ^{2} - 1 ) ^{3}}

(t^{2} - 1)^{3} = t^{6} - 3 t^{4} + 3 t^{2} - 1

= \frac{- 2 t ^{5} - 2 t}{( t ^{4} + 2 t ^{2} + 1 ) ( t ^{6} - 3 t ^{4} + 3 t ^{2} - 1 )}

Expand

(t^{4} + 2 t^{2} + 1) (t^{6} - 3 t^{4} + 3 t^{2} - 1) : t^{10} - t^{8} - 2 t^{6} + 2 t^{4} + t^{2} - 1

= \frac{- 2 t ^{5} - 2 t}{t ^{10} - t ^{8} - 2 t ^{6} + 2 t ^{4} + t ^{2} - 1}

f (3 x^{2} + 4 x - 6) d x

\frac{1}{( y - 1 ) ^{2}}

\frac{4 ( 285 - x )}{9}

\frac{4 x - 9}{x ( x + 3 )}

\frac{2 x - 2}{( 5 - 2 x + x ^{2} ) ^{4}}

What is expand (-2t^5-2t)/((t^2+1)^2(t^2-1)^3) ?
The solution to expand (-2t^5-2t)/((t^2+1)^2(t^2-1)^3) is (-2t^5-2t)/(t^{10)-t^8-2t^6+2t^4+t^2-1}