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
