Use the binomial theorem to expand How can we expand this using the binomial theorem?
$(x^2 + 1/x)^7$
 A: One way is to note that our expression is $\frac{1}{x^7}(1+x^3)^7$ and use the ordinary binomial expansion of $(1+t)^7$. Then substitute $t=x^3$ and (if it is useful) divide the resulting expression for $(1+x^3)^7$ through by $x^7$ term by term. 
Another way is to use the ordinary expansion of $(a+b)^7$, substitute $a=x^2$ and $b=\frac{1}{x}$, and simplify. 
A: Let $a = x^{2}$ and $b = \frac{1}{x}$.  The binomial is written as $(a + b)^{7}$.  Apply Binomial Theorem, so we have:
$$(a + b)^{7} = \dbinom{7}{0} a^{7}b^{7 - 7} + \dbinom{7}{1} a^{6}b^{7 - 6} + \dbinom{7}{2} a^{5}b^{7 - 5} + \dbinom{7}{3} a^{4}b^{7 - 4} + \dbinom{7}{4} a^{3}b^{7 - 3} + \dbinom{7}{5}a^{2}b^{7 - 2} + \dbinom{7}{6}a^{1}b^{7 - 1} + \dbinom{7}{7}a^{0}b^{7 - 0}$$
$$= a^{7} + 7a^{6}b + 21a^{5}b^{2} + 35a^{4}b^{3} + 35a^{3}b^{4} + 21a^{2}b^{5} + 7ab^{6} + b^7$$
Substitute back with $a = x^{2}$ and $b = \frac{1}{x}$, so you get the following answer:
$$(x^{2})^{7} + 7(x^{2})^{6}(\frac{1}{x}) + 21(x^{2})^{5}(\frac{1}{x})^{2} + 35(x^{2})^{4}(\frac{1}{x})^{3} + 35(x^{2})^{3}(\frac{1}{x})^{4} + 21(x^{2})^{2}(\frac{1}{x})^{5} + 7(x^{2})(\frac{1}{x})^{6} + (\frac{1}{x})^7$$
$$= x^{14} + 7x^{11} + 21x^{8} + 35x^{5} + \frac{7}{x^{4}} + 35x^{2} + \frac{21}{x} + \frac{1}{x^7}$$
