Question

Simplify the given expression, using only positive exponents. then complete the statements that follow. [ (x2y3)−1 (x−2y2z)2 ] 2 , x ≠ 0, y ≠ 0, z ≠ 0.

Answer 1

[(x**-2) y**2 .z)**2]**2        (x**-8)(y**8)(z**4)
----------------------- =        ----------------------------- 
[x**2 . y**3]**2                   (x**4) (y**6)


     (y**8)(z**4)            (y**2)*(z**4)
    --------------------- = -------------
    (x**12) (y**6)             (x**12)

Answer 2

Answer:

$x^{-12}y^{2}z^{4}$

Step-by-step explanation:

We have to simplify the given expression given as

$[(x^{2}y^{3})^{-1}.(x^{-2}y^{2}z)^{2}]^{2}$

$[x^{-2}.y^{-3}.x^{-4}.y^{4}.z^{2}]^{2}$

$[x^{-2-4}.y^{-3+4}.z^{2}]^{2}=[x^{-6}.y^{1}.z^{2}]^{2}=x^{(-6-6)}.y^{1+1}.z^{2+2}=x^{-12}y^{2}z^{4}$

Therefore simplification of the given expression gives $x^{-12}y^{2}z^{4}$