Question

Expand $(8-9x^3+18x^4{)}^{\frac{4}{3}}$ as far as $x^8$.

Answer 1

$(8-9x^3+18x^4{)}^{\frac{4}{3}}=16{(1-\frac{9}{8}{x}^3+\frac{9}{4}{x}^4)}^{\frac{4}{3}}$

$=1-\frac{4}{3}(\frac{9}{8}{x}^3-\frac{9}{4}{x}^4)+\frac{\left(\frac{4}{3}\right)\left(\frac{1}{3}\right)}{1\cdot 2}{(\frac{9}{8}{x}^3-\frac{9}{4}{x}^4)}^2\dots \dots$

$=1-\frac{3}{2}{x}^3+3x^4+\frac{9}{32}(x^3-2x^4{)}^2+\dots$

$=1-\frac{3}{2}{x}^3+3x^4+\frac{9}{32}{x}^6-\frac{9}{8}{x}^7+\frac{9}{8}{x}^8$

Multiplying by 16, we get the final equation as

$=16-24x^3+48x^4+\frac{9}{2}{x}^6-18x^7+18x^8$