Respuesta :

We are asked to write in its simplest form the expression :

[tex](27a^{-9})^{(-\frac{2}{3})}[/tex]

So we start by writing the most external power (-2/3) in root form. Recall that a denominator in the exponent's fraction is the index of the radical that has to be used. on the other hand, the numerator of the exponents fraction is a power. Also recall that when there is a NEGATIVE sign in the exponents such implies that the algebraic expression to which it is applied flips to its RECIPROCAL.

Then starting with the negative in the a^-9 we get:

[tex](\frac{27}{a^9})^{(-\frac{2}{3})}[/tex]

and now using the negative sign in the external exponents, we get the reciprocal of the fractional expression as shown below:

[tex](\frac{27}{a^9})^{(-\frac{2}{3})}=(\frac{a^9}{27})^{(\frac{2}{3})}[/tex]

Now we apply the radical form we discussed before:

[tex](\frac{a^9}{27})^{(\frac{2}{3})}=(\sqrt[3]{(\frac{a^9}{27}})^{})^2[/tex]

Now we use the fact that a^9 and 27 are perfect cubes, in order to cancel the cubic root:

[tex]\sqrt[3]{(\frac{a^9}{27}})=\frac{\sqrt[3]{a^{3\cdot3}}}{\sqrt[3]{3^3}}=\frac{a^3}{3}[/tex]

and finally, raise this expression to the power 2 :

[tex](\frac{a^3}{3})^2=\frac{a^6}{9}[/tex]

Therefore a^6/9 is the simplest form of the algebraic expression that was given

RELAXING NOICE
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