Question

Expand to 4 terms the following expressions :
$(8+12a{)}^{\frac{2}{3}}$

Answer 1

Whatever be value of $n$, negative integer or positive or negative fraction,

$(1+x{)}^n=1+\frac{n}{1!}x+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...+...+\frac{n(n-1)(n-2)...(n-r+1)}{r!}{x}^r+...+$ to $\infty$

where, $|x|<1$ i.e. $-1<x<1.$

Now,

$(8+12a{)}^{\frac{2}{3}}=8^{\frac{2}{3}}{(1+\frac{12a}{8})}^{\frac{2}{3}}=4{(1+\frac{3a}{2})}^{\frac{2}{3}}$

$=4[1+\frac{2}{3}\times \frac{3a}{2}+\frac{\frac{2}{3}(\frac{2}{3}-1)}{1.2}{\left(\frac{3a}{2}\right)}^2+\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{\mathrm{1.2.3}}{\left(\frac{3a}{2}\right)}^3+...............]$

$=4[1+a-\frac{a^2}{4}+\frac{a^3}{6}+........]$