Question

Please help

Answer 1

I'm thinking its \[9a^2\]

Answer 2

well the power of 2/3 operates on both the 27 and a^-3

so rewrite 27 as 3^3

then you have

\[(3^3)^{-\frac{2}{3}}\times (a^{-3})^{-\frac{2}{3}}\]

the power of power rule says multiply the powers....

I'll let you finish it... and your answer is half right...

Answer 3

ok so\[\frac{ 1 }{ 9a^2 }\]

Answer 4

@campbell_st

Answer 5

not quite, half right again

it simplifies to

\[3^{-2} a^2 = \frac{a^2}{9}\]

Answer 6

or no it would be \[\frac{ a^2 }{ 9 }\]

Answer 7

yeah I noticed where I messed up thanks so much