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

The solution to expand (1+(3i)/n)^3 is (n^2-27)/(n^2)+(9n^2-27)/(n^3)i

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expand (1+(3i)/n)^3

expand

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

\frac{n ^{2} - 27}{n ^{2}} + \frac{9 n ^{2} - 27}{n ^{3}} i

Solution steps

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

Apply Perfect Cube Formula:

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

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

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

1^{3} = 1

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

3 \cdot 1 \cdot (\frac{3 i}{n})^{2} = - 3 \cdot \frac{9}{n ^{2}}

(\frac{3 i}{n})^{3} = - \frac{27 i}{n ^{3}}

= 1 + 3 \cdot \frac{3 i}{n} - 3 \cdot \frac{9}{n ^{2}} - \frac{27 i}{n ^{3}}

Simplify

1 + 3 \cdot \frac{3 i}{n} - 3 \cdot \frac{9}{n ^{2}} - \frac{27 i}{n ^{3}} : \frac{n ^{2} - 27}{n ^{2}} + \frac{9 n ^{2} - 27}{n ^{3}} i

= \frac{n ^{2} - 27}{n ^{2}} + \frac{9 n ^{2} - 27}{n ^{3}} i

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

(x^{2})^{3}

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

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

(sec (θ) + tan (θ))^{2}

What is expand (1+(3i)/n)^3 ?
The solution to expand (1+(3i)/n)^3 is (n^2-27)/(n^2)+(9n^2-27)/(n^3)i