The expression $\frac{1}{(x^2 + \frac{1}{x})^{4/3}}$ can be expanded by the binomial theorem if:

The fourth term in the expansion of $(1 - 2x)^{3/2}$ is:

$\frac{1}{8} - \frac{7}{8 \times 12} + \frac{7 \times 10}{8 \times 12 \times 16} - \ldots =$

If the value of $x$ is so small that $x^2$ and higher powers can be neglected,then $\frac{\sqrt{1 + x} + \sqrt[3]{(1 - x)^2}}{1 + x + \sqrt{1 + x}}$ is equal to

If $T_4$ represents the $4^{th}$ term in the expansion of $\left(5x + \frac{7}{x}\right)^{-3/2}$ and $x \notin \left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$,then $\left(x^7 \sqrt{5x}\right) T_4 =$

Find the binomial expansion of $\frac{1}{\sqrt[3]{6 - 3x}}$.