Question 948626
1.

{{{(1/25)^(3/2)}}}-> exponent {{{3/2}}} means square root of a number to the power of {{{3}}}

so, you have

{{{(1/25)^(3/2)}}}

={{{sqrt((1/25)^3))}}}

={{{sqrt(1^3)/sqrt(25^3)}}}

={{{1/sqrt((5^2)^3)}}}

={{{1/sqrt(5^2*5^2*5^2)}}}

={{{1/(5*5*5)}}}

={{{1/125}}}

in decimal form: {{{0.008}}}


2.

{{{64^-2/3}}} here you have negative exponent{{{-(2/3)}}} and to get positive exponent write inverse: {{{-(2/3)=1/(2/3)}}}

so, you have

{{{64^-2/3}}}

={{{1/(64^(2/3))}}}

={{{1/(root(3,64^2))}}}...you can write {{{64^2}}} as {{{8^2*8^2}}},then {{{(2^2*2^2*2^2)*(2^2*2^2*2^2)=(2^2)^3(2^2)^3=4^3*4^3}}}

={{{1/(root(3,(4^3*4^3)))}}}

={{{1/(4*4)}}}

={{{1/16}}}

or as decimal {{{0.0625}}}