What is expand ((x+1)(x-1)(x-2)(x-3))/6 ?

The solution to expand ((x+1)(x-1)(x-2)(x-3))/6 is (x^4)/6-(5x^3)/6+(5x^2)/6+(5x)/6-1

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expand ((x+1)(x-1)(x-2)(x-3))/6

expand

\frac{( x + 1 ) ( x - 1 ) ( x - 2 ) ( x - 3 )}{6}

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

Solution steps

\frac{( x + 1 ) ( x - 1 ) ( x - 2 ) ( x - 3 )}{6}

Expand

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

= \frac{x ^{4} - 5 x ^{3} + 5 x ^{2} + 5 x - 6}{6}

Apply the fraction rule:

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

\frac{x ^{4} - 5 x ^{3} + 5 x ^{2} + 5 x - 6}{6} = \frac{x ^{4}}{6} - \frac{5 x ^{3}}{6} + \frac{5 x ^{2}}{6} + \frac{5 x}{6} - \frac{6}{6}

= \frac{x ^{4}}{6} - \frac{5 x ^{3}}{6} + \frac{5 x ^{2}}{6} + \frac{5 x}{6} - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

\frac{6}{6} = 1

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

2 (x_{1} - x_{2} - x_{3})^{2}

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

((2 + cos (u) sin (u)))^{2}

\frac{10 x + 9}{3 ( x + 1 ) ( 2 x + 3 )}

\frac{y ^{2} + x ^{2}}{( y + x ) ^{2} ( x - y ) ^{2}}

What is expand ((x+1)(x-1)(x-2)(x-3))/6 ?
The solution to expand ((x+1)(x-1)(x-2)(x-3))/6 is (x^4)/6-(5x^3)/6+(5x^2)/6+(5x)/6-1