How do you simplify #((4^0c^2d^3f)/(2c^-4d^-5))^-3#?

Question

3238 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-07T01:53:38

#8/(c^18d^24f^3)#

So using the multiplication exponent rule it would simplify to

#((1/64)^0c^-6d^-9f^-3)/(-1/8c^12d^15)#

Using Exponent rules again #(1/64)^0# becomes 1

#(c^-6d^-9f^-3)/(1/8c^12d^15)#

Using the subtraction exponent rule (When you subtract exponents due to division) it simplifies to this:

#(1)/(1/8c^18d^24f^3)#

Simplifying it further by getting rid of the fraction gives:

#8/(c^18d^24f^3)#

Answer 2 · 2017-09-05T06:06:31

#8/(c^18d^24f^3)#

#((4^0c^2d^3f)/(2c^-4d^-5))^-3#

Recall the law of indices: #(a/b)^-m = (b/a)^m#

Start by getting rid of the negative indices:

#((4^0c^2d^3f)/(2c^-4d^-5))^color(red)(-3) = ((2c^-4d^-5)/(4^0c^2d^3f))^color(red)(3)" "larr# invert the fraction

Recall the law of indices: #x^(-m) = 1/x^m#

#((2color(blue)(c^-4d^-5))/(4^0c^2d^3f))^3 = ((2)/(4^0c^2d^3fxxcolor(blue)(c^4d^5)))^3#

Recall the laws: #x^0=1 and x^m xx x^n = x^(m+n)#

# ((2)/(cancel(4^0)^1c^2d^3fc^4d^5))^3 = ((2)/(c^6d^8f))^3#

Recall the law: #(x^m)^n =x^(mn)#

#((2)/(c^6d^8f))^3 = 8/(c^18d^24f^3)" "larr# multiply the indices