What is expand ((4x-1)^4)/(16) ?

The solution to expand ((4x-1)^4)/(16) is 16x^4-16x^3+6x^2-x+1/16

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expand ((4x-1)^4)/(16)

expand

\frac{( 4 x - 1 ) ^{4}}{16}

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

Solution steps

\frac{( 4 x - 1 ) ^{4}}{16}

(4 x - 1)^{4} = 256 x^{4} - 256 x^{3} + 96 x^{2} - 16 x + 1

= \frac{256 x ^{4} - 256 x ^{3} + 96 x ^{2} - 16 x + 1}{16}

Apply the fraction rule:

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

\frac{256 x ^{4} - 256 x ^{3} + 96 x ^{2} - 16 x + 1}{16} = \frac{256 x ^{4}}{16} - \frac{256 x ^{3}}{16} + \frac{96 x ^{2}}{16} - \frac{16 x}{16} + \frac{1}{16}

= \frac{256 x ^{4}}{16} - \frac{256 x ^{3}}{16} + \frac{96 x ^{2}}{16} - \frac{16 x}{16} + \frac{1}{16}

\frac{256 x ^{4}}{16} = 16 x^{4}

\frac{256 x ^{3}}{16} = 16 x^{3}

\frac{96 x ^{2}}{16} = 6 x^{2}

\frac{16 x}{16} = x

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

cos (2 (π - θ))

\frac{1}{2} (u - \frac{1}{2} sin (2 u))

\frac{\frac{d}{d x} ( ( 6 x ^{2} - 3 ) ^{4} )}{( 2 x ^{3} + 8 ) ^{5}}

(1 - sin (x))^{2} + (cos (x))^{2}

\exists x (x^{3} = - 1)

What is expand ((4x-1)^4)/(16) ?
The solution to expand ((4x-1)^4)/(16) is 16x^4-16x^3+6x^2-x+1/16