What is (-1+(2i)/n)^3 ?

The solution to (-1+(2i)/n)^3 is -(n^2-12)/(n^2)+(6n^2-8)/(n^3)i

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simplify (-1+(2i)/n)^3

simplify

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

- \frac{n ^{2} - 12}{n ^{2}} + \frac{6 n ^{2} - 8}{n ^{3}} i

Solution steps

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

Apply Perfect Cube Formula:

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

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

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

(- 1)^{3} = - 1

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

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

(\frac{2 i}{n})^{3} = - \frac{8 i}{n ^{3}}

= - 1 + 3 \cdot \frac{2 i}{n} + 3 \cdot \frac{4}{n ^{2}} - \frac{8 i}{n ^{3}}

Simplify

- 1 + 3 \cdot \frac{2 i}{n} + 3 \cdot \frac{4}{n ^{2}} - \frac{8 i}{n ^{3}} : - \frac{n ^{2} - 12}{n ^{2}} + \frac{6 n ^{2} - 8}{n ^{3}} i

= - \frac{n ^{2} - 12}{n ^{2}} + \frac{6 n ^{2} - 8}{n ^{3}} i

4 r^{2} = 100

12 x^{6} - 3 x^{2}

- 4 x^{3} \cdot 2 y^{- 2} \cdot 5 y^{5} \cdot x^{- 8}

1 0^{2}

\frac{3 a - 36}{12 b - ab}

What is (-1+(2i)/n)^3 ?
The solution to (-1+(2i)/n)^3 is -(n^2-12)/(n^2)+(6n^2-8)/(n^3)i