How do you express (3x^2)^-3(3x2)−3 with positive exponents?

2 Answers
Shwetank Mauria
Sep 11, 2016

(3x^2)^(-3)=1/(27x^6)(3x2)−3=127x6

Explanation:

We can use the identities a^(-m)=1/a^ma−m=1am, (a^m)^n=a^(mn)(am)n=amn and (ab)^m=a^mb^m(ab)m=ambm

Hence, (3x^2)^(-3)(3x2)−3

= 1/(3x^2)^31(3x2)3

= 1/(3^3*(x^2)^3)133⋅(x2)3

= 1/(27*x^(2xx3))127⋅x2×3

= 1/(27x^6127x6

Tony B
Sep 18, 2016

1/(27x^6)127x6

Explanation:

Method Example:

Suppose we had t^(-3)t−3. This is another way of writing 1/(t^3)1t3

Suppose we had 1/(t^-3)1t−3 this is another way of writing t^3t3

So the negative power (index) has the effect of moving the value to the other side of the horizontal line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:" "(3x^2)^-3 (3x2)−3

The index -3−3 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(3x2)3 → 133×x2×x2×x2 → 133x2×3

=1/(27x^6)=127x6

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