Solve for \( x: \frac{6}{x+2}-\frac{1}{-2}=\frac{6}{x} \) (no fractions) (in the form \( a x^{2}+b x+c=0 \) ) (type: " \( x= \) sol is not applicable" if one of the solutions is not applicable)

(4) \( (x+3)^{3}=343 \)

Use the power of a power property to simplify: \( (2y^{5})^{2} \)

\[ h(x)=x^{3}+a x^{2}+b x+c \] The function \( h \) is defined above, where \( a, b \), and \( c \) are integer constants. If the zeros of the function are \( -5,6 \), and 7 , what is the value of \( c \) ?

\( \frac{\mathbf{2}^{2004}-\mathbf{2}^{2002}}{2^{2001}}=? \) \( \begin{array}{lllll}\text { A) } 2 & \text { B) } 4 & \text { C) } 6 & \text { D) } 8 & \text { E) }\end{array} \)

8. Expresa con exponente fraccionario y simplifica si es posible. \( \begin{array}{llll}\text { a) } \sqrt{\left(x^{4} y^{6}\right)^{2}} & \text { b) } \sqrt[3]{x^{2}+y^{2}} & \text { c) } \sqrt[5]{(3 y)^{7}} & \text { d) } \sqrt[5]{\frac{2}{x^{3}}}\end{array} \)