What is expand ((x^2y^2-1)^4)/(x^4y^4) ?

The solution to expand ((x^2y^2-1)^4)/(x^4y^4) is x^4y^4-4x^2y^2+6-4/(x^2y^2)+1/(x^4y^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ((x^2y^2-1)^4)/(x^4y^4)

expand

\frac{( x ^{2} y ^{2} - 1 ) ^{4}}{x ^{4} y ^{4}}

x^{4} y^{4} - 4 x^{2} y^{2} + 6 - \frac{4}{x ^{2} y ^{2}} + \frac{1}{x ^{4} y ^{4}}

Solution steps

\frac{( x ^{2} y ^{2} - 1 ) ^{4}}{x ^{4} y ^{4}}

(x^{2} y^{2} - 1)^{4} = x^{8} y^{8} - 4 x^{6} y^{6} + 6 x^{4} y^{4} - 4 x^{2} y^{2} + 1

= \frac{x ^{8} y ^{8} - 4 x ^{6} y ^{6} + 6 x ^{4} y ^{4} - 4 x ^{2} y ^{2} + 1}{x ^{4} y ^{4}}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{x ^{8} y ^{8} - 4 x ^{6} y ^{6} + 6 x ^{4} y ^{4} - 4 x ^{2} y ^{2} + 1}{x ^{4} y ^{4}} = \frac{x ^{8} y ^{8}}{x ^{4} y ^{4}} - \frac{4 x ^{6} y ^{6}}{x ^{4} y ^{4}} + \frac{6 x ^{4} y ^{4}}{x ^{4} y ^{4}} - \frac{4 x ^{2} y ^{2}}{x ^{4} y ^{4}} + \frac{1}{x ^{4} y ^{4}}

= \frac{x ^{8} y ^{8}}{x ^{4} y ^{4}} - \frac{4 x ^{6} y ^{6}}{x ^{4} y ^{4}} + \frac{6 x ^{4} y ^{4}}{x ^{4} y ^{4}} - \frac{4 x ^{2} y ^{2}}{x ^{4} y ^{4}} + \frac{1}{x ^{4} y ^{4}}

\frac{x ^{8} y ^{8}}{x ^{4} y ^{4}} = x^{4} y^{4}

\frac{4 x ^{6} y ^{6}}{x ^{4} y ^{4}} = 4 x^{2} y^{2}

\frac{6 x ^{4} y ^{4}}{x ^{4} y ^{4}} = 6

\frac{4 x ^{2} y ^{2}}{x ^{4} y ^{4}} = \frac{4}{x ^{2} y ^{2}}

= x^{4} y^{4} - 4 x^{2} y^{2} + 6 - \frac{4}{x ^{2} y ^{2}} + \frac{1}{x ^{4} y ^{4}}

- \frac{4 - 2 s}{s ( s + 2 )}

\frac{x ^{3} - 2}{( 4 x + 3 ) ( 6 x - 2 )}

(1 - x)^{4}

(x^{2} + 7 - 4) (x^{2} + 7 + 4)

(8 - 2 ln (x)) x

What is expand ((x^2y^2-1)^4)/(x^4y^4) ?
The solution to expand ((x^2y^2-1)^4)/(x^4y^4) is x^4y^4-4x^2y^2+6-4/(x^2y^2)+1/(x^4y^4)