Expand the following expression:
$(1 - 2 x)^{3}$

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Solution

Using binomial theorem for the expansion of $(1 - 2 x)^{5}$ , we have

$(1 - 2 x)^{5} = 5_{C_{0}} + 5_{C_{1}} (- 2 x) + 5_{C_{2}} (- 2 x)^{2} + 5_{C_{3}} (- 2 x)^{3} + 5_{C_{4}} (- 2 x)^{4} + 5_{C_{5}} (- 2 x)^{5}$

$= 1 + 5 (- 2 x) + 10 (- 2 x)^{2} + 10 (- 2 x)^{3} + 5 (- 2 x)^{4} + (- 2 x)^{5}$

$= 1 - 10 x + 40 x^{2} - 80 x^{3} + 80 x^{4} - 32 x^{5}$

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Similar questions

(1 + 2 x)^{3}

\frac{4}{3} x^{7} - 3 x^{5} + 2 x^{3} + 1

x

2^{x^{3}} - 1 + \frac{1}{2^{x^{3}} + 2}

\frac{1 - x^{2}}{2 x^{3} - 1} \div \frac{1 - x}{2 x^{2}}

(1)

x = \infty

(2)

x = 0

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

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