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

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

1^{3} = 1

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

3 \cdot 1 \cdot (\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{1}{2} cos^{3} (t) (1 - tan^{4} (t) + sec^{4} (t))

cos (x) (2 sin (x) - 1)

((V - \frac{k \cdot c}{t \cdot s + 1}) A) F

x^{\frac{4}{5}} (8 x - 32)^{2}

\frac{x - 1}{( x + 1 ) ( x - 3 )}

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