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

The solution to expand ((x-1)x^2(x+1)^2)/4 is (x^5)/4+(x^4)/4-(x^3)/4-(x^2)/4

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

expand

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

\frac{x ^{5}}{4} + \frac{x ^{4}}{4} - \frac{x ^{3}}{4} - \frac{x ^{2}}{4}

Solution steps

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

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

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

Expand

(x - 1) x^{2} (x^{2} + 2 x + 1) : x^{5} + x^{4} - x^{3} - x^{2}

= \frac{x ^{5} + x ^{4} - x ^{3} - x ^{2}}{4}

Apply the fraction rule:

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

\frac{x ^{5} + x ^{4} - x ^{3} - x ^{2}}{4} = \frac{x ^{5}}{4} + \frac{x ^{4}}{4} - \frac{x ^{3}}{4} - \frac{x ^{2}}{4}

= \frac{x ^{5}}{4} + \frac{x ^{4}}{4} - \frac{x ^{3}}{4} - \frac{x ^{2}}{4}

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

\frac{( x ^{3} + 6 ) ^{2}}{225}

\frac{8 x ^{2} - 6 x}{( 8 x - 3 ) ^{2}}

4 (5 x - 7 = 10 x + 2)

\frac{1}{s ( s + 1 ) ( s + 5 ) ( s + 10 )}

What is expand ((x-1)x^2(x+1)^2)/4 ?
The solution to expand ((x-1)x^2(x+1)^2)/4 is (x^5)/4+(x^4)/4-(x^3)/4-(x^2)/4