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

The solution to expand ((y-1)/2)^4 is (y^4}{16}-\frac{y^3)/4+(3y^2)/8-y/4+1/16

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

expand

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

\frac{y ^{4}}{16} - \frac{y ^{3}}{4} + \frac{3 y ^{2}}{8} - \frac{y}{4} + \frac{1}{16}

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

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

2^{4} = 16

= \frac{y ^{4} - 4 y ^{3} + 6 y ^{2} - 4 y + 1}{16}

Apply the fraction rule:

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

\frac{y ^{4} - 4 y ^{3} + 6 y ^{2} - 4 y + 1}{16} = \frac{y ^{4}}{16} - \frac{4 y ^{3}}{16} + \frac{6 y ^{2}}{16} - \frac{4 y}{16} + \frac{1}{16}

= \frac{y ^{4}}{16} - \frac{4 y ^{3}}{16} + \frac{6 y ^{2}}{16} - \frac{4 y}{16} + \frac{1}{16}

Cancel

\frac{4 y ^{3}}{16} : \frac{y ^{3}}{4}

= \frac{y ^{4}}{16} - \frac{y ^{3}}{4} + \frac{6 y ^{2}}{16} - \frac{4 y}{16} + \frac{1}{16}

Cancel

\frac{6 y ^{2}}{16} : \frac{3 y ^{2}}{8}

= \frac{y ^{4}}{16} - \frac{y ^{3}}{4} + \frac{3 y ^{2}}{8} - \frac{4 y}{16} + \frac{1}{16}

Cancel

\frac{4 y}{16} : \frac{y}{4}

= \frac{y ^{4}}{16} - \frac{y ^{3}}{4} + \frac{3 y ^{2}}{8} - \frac{y}{4} + \frac{1}{16}

\frac{4 ( 3 ^{n} - 1 )}{3}

(3 x^{2} + 5) ln (2)

\frac{p}{( ( p ^{2} + 4 ) ( p ^{2} + 9 ) )}

t^{b} a + \frac{d}{d x} (3 x^{2} (7 - x)^{5})

((A x + B) e^{x}) x

What is expand ((y-1)/2)^4 ?
The solution to expand ((y-1)/2)^4 is (y^4}{16}-\frac{y^3)/4+(3y^2)/8-y/4+1/16