Hey guys, need help here

Assuming [tex]a[/tex] is a non-negative real number...
Raise both sides to the power 2 :
[tex]a^{3/4} = 8 \implies \left(a^{3/4}\right)^2 = 8^2 \implies a^{3/2} = 64[/tex]
which follows from the exponent product property, [tex](x^y)^z=x^{yz}[/tex]. In particular, 3/4 × 2 = 6/4 = 3/2.
Raise both sides to the power 1/3 :
[tex]\left(a^{3/2}\right)^{1/3} = 64^{1/3} \implies a^{1/2} = 4[/tex]
where we use the same property as before, and 4³ = 64. This time, 3/2 × 1/3 = 3/6 = 1/2.
Take the reciprocal of both sides. This negates the exponent, so we end up with
[tex]\dfrac1{a^{1/2}} = \dfrac14 \implies \boxed{a^{-1/2} = \dfrac14}[/tex]
Of course, you could also solve for [tex]a[/tex] immediately by raising both sides of the original equation to the power 4/3. We have 4/3 × 3/4 = 12/12 = 1, so
[tex]a^{3/4} = 8 \implies \left(a^{3/4}\right)^{4/3} = 8^{4/3} \implies a = 8^{4/3}[/tex]
Now 2³ = 8, so [tex]8^{4/3} = \left(8^{1/3}\right)^4 = 2^4 = 16[/tex], and since 4² = 16, it follows that
[tex]a = 16 \implies a^{1/2} = \sqrt{16} = 4 \implies a^{-1/2} = \dfrac14[/tex]