Help wih math problems

Apply exponent rules to simplify:
(-2x^2y^-3z)^3 (only positive exponents should be in answer)
(2x^-3)(2x^4) (only positive exponents should be in answer)

Perform operation and simplify :
2x^2+3x-8-(x^2+7)
(x+2)^3
3x^2/x^2-16 ÷ 9x/x+4

xxxxxxxxxxxxxxxxxxxxxxxxx divided by who cares

order of operations do the -8 (x^2+7)
it will become -8x^2+(-56)
so now you have 2x^2+3x -8x^2+(-56) just group all like terms and the you get
10x^2+3x+(-56) when you add variables with exponents the exponents are not added or subtracted that just lets you know there the same and you can combine them.
(x+2)^3 with this you raising the x^3 and the 2 to 6 so its x^3+6
3x^2/x^2-16 ÷ 9x/x+4 since its a division problem you flip the second one and it becomes x+4/9x
then you cross multiply and get x^3-16x+4x^2-64/21x^3....thats as far as i go hope it helps!

These are pretty ambiguous because of the plain-text format. Reading them as is, assuming that there are no parentheses unless indicated, and following PEMDAS, here's how to do them:

(-2x^2y^-3z)^3 = (-2x^(-6yz))^3 = (-2)^3(x^(-9yz)) = -8/(x^(9yz))

(2x^-3)(2x^4)= (4x^4-3) = 4x [If it's [(2x)^-3][(2x)^4)], you get (2x)^4-3 = 2x]

2x^2 + 3x - 8 - (x^2 + 7) = x^2 + 3x - 15

(x+2)^3 = (x+2)(x+2)(x+2) = (x^2 + 4x + 4)(x+2) = (x^3 + 4x^2 + 4x + 2x^2 + 8x + 8) = x^3 + 6x^2 + 12x +8

((3x^2/x^2) - 16) / ((9x/x) + 4) = (3 - 16) / (9 + 4) = -13/13 = -1

 = [ - 2 • x² • y‾³ • z ]³
 = - 8 • (x^6) • (y^-9) • z³
 = - 8 • (x^6) • z³ ⁄ (y^9)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~…

   = 2 • (x^-3) •  2 • (x^4)
   = 4 • (x¹)
   = 4x

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~…

You get the idea.

(-2x^2y^-3z)^3
=(-2x^6y^-9z^3)
=(-2x^6z^3) / (y^9)

(2x^-3)(2x^4)
= 4x

2x^2 + 3x - 8 - (x^2 + 7)
= 2x^2 + 3x - 8 - x^2 - 7
= x^2 + 3x - 15

(x + 2)^3
= (x + 2)(x + 2)(x + 2)
= (x^2 + 4x + 4)(x + 2)
= x^3 + 2x^2 + 4x^2 + 8x + 4x + 8
= x^3 + 6x^2 + 12x + 8

(3x^2/x^2-16) / (9x/x+4)
= (3x^2/x^2-16)(x+4/9x)
= (3x^2 * x+4) / (x^2-16 * 9x)
= (3x^2 * x+4) / [(x - 4)(x + 4)(9x)]
= x / [3(x - 4)]