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

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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

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