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

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

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expand (1+(2i)/n)^3

expand

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

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

1^{3} = 1

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

3 \cdot 1 \cdot (\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

(8 + sin (t))^{2} + cos^{2} (t)

sec (y) (tan (y) - sec (y)) d y

H (8 - t) (sin (t - 1) - sin (t - 4))

f (x) = (3 - e^{x})^{2}

\frac{5 s ^{2} + 4}{( s ^{2} + 17 ) ( s - 1 )}

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