Constant term

It's    60.

Using the binomial theorem,the fifth term is (6)(5)(4)(3)(2x^2)^2(1/x)^4    /  4!

=(6)(5)(4)(3)(4x^4)(1/x^4)     /4!

=(6)(5)(4)(3) (4)       /4!

= (6)(5)(4)(3)  (4)/     (4)(3)(2)

=(6)(5)(4)  /(2)     = (6)(5)(2)  = 30 x 2  = 60

The constant  willl occur at the 5th  term in the binomial expansion of this  =

C(6,4) * (2x^2)^2 * (1/x)^4  =

15 (4x^4) (1/x^4)  =

15* 4   =

60

how do i find the constant term in the expansion of (2x^2-1/x)^6

$\begin{array}{|rcll|} \hline && \left( 2x^2-\frac{1}{x} \right)^6 \\\\ &=& \left( \frac{2x^3-1}{x} \right)^6 \\\\ &=& \frac{(2x^3-1)^6}{x^6} \\\\ &=& \frac{1}{x^6} \cdot (2x^3-1)^6 \\ \hline \end{array}$

The constant  term is at the 5th term in the binomial expansion:

$\begin{array}{|rcll|} \hline && \frac{1}{x^6} \cdot \binom64 \cdot (2x^3)^2 \quad & | \quad \binom64 = \binom{6}{6-4} = \binom62\\ &=& \frac{1}{x^6} \cdot \binom62 \cdot (2x^3)^2 \\ &=& \frac{1}{x^6} \cdot \binom62 \cdot 4x^6 \\ &=& 4 \cdot \binom62 \\ &=& 4 \cdot \frac62\cdot \frac51 \\ &=& 4 \cdot 3\cdot 5 \\ &=& 60 \\ \hline \end{array}$