How do you simplify ${(x^{- 1} y^{- 2} z^{3})}^{- 2} (x^{2} y^{- 4} z^{6})$ $(x^-1y^-2z^3)^-2 (x^2y^-4z^6)$ ?

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Answer 1 · 2016-03-15T00:38:05

Use the following three properties:

$(x*y*z*...)^a=x^a*y^a*z^a*...$

$(x^a)^b=x^(a*b)$

$x^a*x^b=x^(a+b)$

Answer is:

$x^4y^0z^0$

$(x^-1y^-2z^3)^-2(x^2y^-4z^6)$

$(x^-1)^-2(y^-2)^-2(z^3)^-2x^2y^-4z^6$

$x^(-1*(-2))y^(-2*(-2))z^(3*(-2))x^2y^-4z^6$

$x^2y^4z^-6x^2y^-4z^6$

$x^(2+2)y^(4+(-4))z^(-6+6)$

$x^4y^0z^0$

Note: don't say that $y^0=1$ and $z^0=1$ because that is not true for $y=0$ and $z=0$

If you wish you can use $x^-a=1/x^a$ but it's easier without it.

How do you simplify (x^-1y^-2z^3)^-2 (x^2y^-4z^6)(x−1y−2z3)−2(x2y−4z6)(x^-1y^-2z^3)^-2 (x^2y^-4z^6)?