How do you simplify ((4^0c^2d^3f)/(2c^-4d^-5))^-3(40c2d3f2c4d5)3?

2 Answers
Jun 7, 2017

8/(c^18d^24f^3)8c18d24f3

Explanation:

So using the multiplication exponent rule it would simplify to

((1/64)^0c^-6d^-9f^-3)/(-1/8c^12d^15)(164)0c6d9f318c12d15

Using Exponent rules again (1/64)^0(164)0 becomes 1

(c^-6d^-9f^-3)/(1/8c^12d^15)c6d9f318c12d15

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

(1)/(1/8c^18d^24f^3)118c18d24f3

Simplifying it further by getting rid of the fraction gives:

8/(c^18d^24f^3)8c18d24f3

Sep 5, 2017

8/(c^18d^24f^3)8c18d24f3

Explanation:

((4^0c^2d^3f)/(2c^-4d^-5))^-3(40c2d3f2c4d5)3

Recall the law of indices: (a/b)^-m = (b/a)^m(ab)m=(ba)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(40c2d3f2c4d5)3=(2c4d540c2d3f)3 invert the fraction

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

((2color(blue)(c^-4d^-5))/(4^0c^2d^3f))^3 = ((2)/(4^0c^2d^3fxxcolor(blue)(c^4d^5)))^3(2c4d540c2d3f)3=(240c2d3f×c4d5)3

Recall the laws: x^0=1 and x^m xx x^n = x^(m+n)x0=1andxm×xn=xm+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