See both the pics...

I hope it helps...

thanku

Verified answer

[tex]\bf{Given~expression}[/tex]

[tex]\large\sf{\longrightarrow (64)^{\frac{-1}{6}} }[/tex]

Since the power is non-integral, we try to get rid of that and have an integral exponent

[tex]\large\sf{\longrightarrow (64)^{\frac{-1}{6}} }[/tex]

[tex]\large\sf{\longrightarrow (2^6)^{\frac{-1}{6}} }[/tex]

Using the law -

[tex]\dashrightarrow \large\underline{\boxed{\bf{ (a^m)^n = a^{m.n}}}}♣[/tex]

[tex]\large\sf{\longrightarrow 2^{6 × \frac{-1}{6}} }[/tex]

[tex]\large\sf{\longrightarrow 2^{-1} }[/tex]

So, we found that

[tex]\large\bf{\dashrightarrow (64)^{\frac{-1}{6}} = 2^{-1} }[/tex]

Now, to get rid of the negative exponent, we use the law mentioned below-

[tex]\dashrightarrow\large \underline{\boxed{\bf{a^{-m} = \dfrac{1}{a^m} }}}♣[/tex]

Using this in our expression we get -

[tex]\large\bf{\mapsto 2^{-1} = \dfrac{1}{2} }[/tex]

Therefore,

[tex]\hookrightarrow \large \boxed{\bf{ (64)^{\frac{-1}{6}} = \dfrac{1}{2} }}[/tex]

Now, if we have to find the reciprocal of [tex]\sf{(64)^{\frac{-1}{6}}}[/tex]

If we have to find the reciprocal of any number [tex]\sf{K}[/tex]

we find it as

[tex]\sf{\longrightarrow Reciprocal~of~'K' = K^{-1} }[/tex]

[tex]\rm{using~this,~we~get}[/tex]

since

[tex]\large\sf{\longrightarrow (64)^{\dfrac{-1}{6}}= 2^{-1} }[/tex]

So,

Inverse of [tex]\sf{2^{-1}}[/tex]

[tex]\large\rm{\leadsto ( 2^{-1})^{-1}}[/tex]

[tex]\large\sf{\longrightarrow 2^{-1×(-1) } }[/tex]

[tex]\large\sf{\longrightarrow 2^{1} }[/tex]

[tex]\large\sf{\longrightarrow 2 }[/tex]

Therefore,

[tex]\hookrightarrow \boxed{\bf{Reciprocal~of~(64)^{\frac{-1}{6}} = 2 }}[/tex]

[tex]\hookrightarrow \boxed{\bf{ (64)^{\frac{-1}{6}} = \dfrac{1}{2} }}[/tex]

[tex]\rule{200pt}{5pt}[/tex]

Laws used -

[tex]\dashrightarrow \large\boxed{\bf{ (a^m)^n = a^{m.n}}}♣[/tex]

[tex]\dashrightarrow\large \boxed{\bf{a^{-m} = \dfrac{1}{a^m} }}♣[/tex]