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Q. Expand $\left(x^2+\frac{3}{x}\right)^4\:,\:x\:\ne \:0$

Binomial Theorem

Solution:

We have ;
$\left(x^{2}+\frac{3}{x}\right)^{4} = \,{}^{4}C_{0}\left(x^{2}\right)^{4}+ \,{}^{4}C_{1}\left(x^{2}\right)^{3}\left(\frac{3}{x}\right)+ \,{}^{4}C_{2}\left(x^{2}\right)^{2}\left(\frac{3}{x}\right)^{2}$
$+ \,{}^{4}C_{3}\left(x^{2}\right)\left(\frac{3}{x}\right)^{3}+ \,{}^{4}C_{4}\left(\frac{3}{x}\right)^{4}$
$= x^{8}+4x^{6}\cdot\frac{3}{x}+6x^{4}\cdot\frac{9}{x^{2}}+4x^{2}\cdot\frac{27}{x^{3}}+\frac{81}{x^{4}}$
$= x^{8}+12x^{5}+54x^{2}+\frac{108}{x}+\frac{81}{x^{4}}$.