What is (-1+(4i)/n)^3 ?

The solution to (-1+(4i)/n)^3 is -(n^2-48)/(n^2)+(12n^2-64)/(n^3)i

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simplify (-1+(4i)/n)^3

simplify

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

- \frac{n ^{2} - 48}{n ^{2}} + \frac{12 n ^{2} - 64}{n ^{3}} i

Solution steps

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

Apply Perfect Cube Formula:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

(- 1 + \frac{4 i}{n})^{3} = (- 1)^{3} + 3 (- 1)^{2} \frac{4 i}{n} + 3 (- 1) (\frac{4 i}{n})^{2} + (\frac{4 i}{n})^{3}

= (- 1)^{3} + 3 (- 1)^{2} \frac{4 i}{n} + 3 (- 1) (\frac{4 i}{n})^{2} + (\frac{4 i}{n})^{3}

(- 1)^{3} = - 1

3 (- 1)^{2} \frac{4 i}{n} = 3 \cdot \frac{4 i}{n}

3 (- 1) (\frac{4 i}{n})^{2} = 3 \cdot \frac{16}{n ^{2}}

(\frac{4 i}{n})^{3} = - \frac{64 i}{n ^{3}}

= - 1 + 3 \cdot \frac{4 i}{n} + 3 \cdot \frac{16}{n ^{2}} - \frac{64 i}{n ^{3}}

Simplify

- 1 + 3 \cdot \frac{4 i}{n} + 3 \cdot \frac{16}{n ^{2}} - \frac{64 i}{n ^{3}} : - \frac{n ^{2} - 48}{n ^{2}} + \frac{12 n ^{2} - 64}{n ^{3}} i

= - \frac{n ^{2} - 48}{n ^{2}} + \frac{12 n ^{2} - 64}{n ^{3}} i

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

(2 sin (x) - 1) (sin (x) + 2)

sin ((2 n - 1) \frac{π}{4})

3 x^{3} (4 cos (x) - x sin (x))

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

What is (-1+(4i)/n)^3 ?
The solution to (-1+(4i)/n)^3 is -(n^2-48)/(n^2)+(12n^2-64)/(n^3)i