Radically Rooted Reciprocals

$\begin{aligned} \left[ \left( \dfrac {1}{3} \right)^{-\frac{4}{3}} \right]^{\frac{9}{2}} \left(9^{-3}\right) & = \left[ \left( 3^{-1} \right)^{-\frac{4}{3}} \right]^{\frac{9}{2}} \left((3^2)^{-3}\right) = \left( 3^{\frac{4}{3}} \right)^{\frac{9}{2}} \left(3^{-6}\right) \\ & = \left( 3^{\frac{4}{3} \times \frac{9}{2}} \right) \left(3^{-6}\right) = \left( 3^{6} \right) \left(3^{-6}\right) = 3^{6-6} = 3^0 = \boxed{1} \end{aligned}$

(1/3)= 3^(-1)

(((3^-1)^-4/3)^9/2) x (3^2)^-3

3^(-1 * -4/3 * 9/2) x 3^(2 * -3)

(3^6) * (3^(-6))

(3^6)/(3^6) = 1

Hence the answer......

Yeah my answer is true...yooooo

-4/3 9/2= -6 3^6 1/3^6 =1

(-4/3)(9/2)=-6 ((1/3)^-6)*(9^-3)=(9^-3)/(3^-6)=(3^6)/(9^3)=729/729=1

First, look at the ranks. I saw two ranks between brackets. So I calculate -4/3 times 9/2 equals to -6. Then, change 1/3 into rank form, that is equals to 3 rank minus one, so I used -1 to be timed with -6, I got 6. So now I have 3 rank 6. 3 rank 6 is equal to 729. Now what's with 9 rank minus 3? Yeah, change it into fraction form, that is equals to one per 9 rank 3. Why no minus? Because it goes inside denominator side. 9 rank 3 is equal to 729, so 9 rank minus 3 is equals to one per 729. I have a same numerator and denominator number, so it goes 729/729, you got 1. The answer is 1.

Use John's Derirative Binomial Expansion Theorem.

=[3^(4/3)]^9/2x(3^-6) =3^[(4/3)x(9/2)]x(3^-6) =(3^6)(3^-6) =3^0 =1

(1/3)^-6x9^-3 =3^6x3^-6=1

Consider ((1/3)^-4/3)^9/2 => (1/3)^-4/3 * 9/2 => (1/3)^-4/2 * 9/3 => (1/3)^-6 => (3)^6 * 9^-3 => (3^2)^3 * 9^-3 => 9^3 * 9^-3 => 9^(3-3) => 9^0 = 1

[(1/3)^-4/3]^9/2 term in the expression can be rewritten as [(1/3)^[(-4/3)X(2/3)]], which simplifies to [(1/3)^-6], which then can be rewritten as 3^6. The 9^-3 term can be rewritten as (3^2)^-3 which simplifies to 3^-6. Therefore, (3^6)X(3^-6)=[3^(6-6)]=3^0=1