How do you express #(3x^2)^-3# with positive exponents?

2 Answers
Shwetank Mauria
Sep 11, 2016

#(3x^2)^(-3)=1/(27x^6)#

Explanation:

We can use the identities #a^(-m)=1/a^m#, #(a^m)^n=a^(mn)# and #(ab)^m=a^mb^m#

Hence, #(3x^2)^(-3)#

= #1/(3x^2)^3#

= #1/(3^3*(x^2)^3)#

= #1/(27*x^(2xx3))#

= #1/(27x^6#

Tony B
Sep 18, 2016

#1/(27x^6)#

Explanation:

Method Example:

Suppose we had #t^(-3)#. This is another way of writing #1/(t^3)#

Suppose we had #1/(t^-3)# this is another way of writing #t^3#

So the negative power (index) has the effect of moving the value to the other side of the horizontal line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "(3x^2)^-3#

The index #-3# is acting on everything inside the brackets. So this is the same as:

#1/(3x^2)^3" "->" "1/(3^3xx x^2 xx x^2 xx x^2)" "->" "1/(3^3x^(2xx3))#

#=1/(27x^6)#

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