化简 (i/n-1)^3

解答

化简

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

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

求解步骤

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

使用完全立方公式:

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

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

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

化简

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

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

化简

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

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

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