Question 73491
{{{(2x^-6y^-4z^0/3^-2x^-2y^-8z^5)^3}}}Start with the original problem
{{{(2^3x^(-6*3)y^(-4*3)z^(0*3)/3^(-2*3)x^(-2*3)y^(-8*3)z^(5*3))}}}Since {{{(x^y)^z=x^(y*z)}}} the problem can be rewritten as this. Remember each term is raised to the 3rd power, so this rule applies to each term.
{{{(8x^(-18)y^(-12)z^(0)/3^(-6)x^(-6)y^(-24)z^(15))}}}Simplify the exponents
{{{(8x^(-18-(-6))y^(-12-(-24))z^(0-15)/3^(-6)x^(0)y^(0)z^(0))}}}Remember to divide variables with exponents, you just subtract the exponents. (eg {{{x^2/x^1=x^(2-1)=x^1}}} Also {{{x^0=1}}}
{{{(8x^(-12))y^(12)z^(-15)/3^(-6))}}}
Finally anything raised to a negative power is inverted (ie {{{x^(-2)=1/x^2}}})
{{{(3^6)(8)y^(12)/x^(12)z^(15))}}}
{{{(729)(8)y^(12)/x^(12)z^(15))}}}
{{{(5832)y^(12)/x^(12)z^(15))}}}Here it is in simplified form