There are two things to consider.
1) Are there any squared values we can 'take out' of the roots?
2) It is considered not good practice to have a root as or in the denominator so we need to change what is left of it into a whole number.
Given: #sqrt(h^3)/sqrt(8)#
Write as: #sqrt(h^2xxh)/(sqrt(2^2xx2))color(white)("ddd")" giving: "color(white)("ddd") (hsqrt(h))/(2sqrt(2)) #
Now we 'get rid of the root in the denominator; multiply by 1 and you do not change the inherent value. However 1 comes in many forms.
#color(green)((hsqrt(h))/(2sqrt(2))color(red)(xx1) color(white)("dddd")-> color(white)("dddd")(hsqrth)/(2sqrt2)color(red)(xxsqrt2/sqrt2) )#
#color(white)("d")color(green)(color(white)("ddddddddddd")->color(white)("dddd")(hsqrt(2h)) /4 )#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Foot note")#
Compare the consequence of #sqrt(h)xxsqrt(2)->sqrt(2h)color(white)("d")# to the example:
#sqrt(4)xxsqrt(9)color(white)("d")=color(white)("d")2xx3=6#
#sqrt(4xx9)color(white)("d")=color(white)("d")sqrt(36)color(white)("d")=color(white)("d")6#
#color(brown)("So " hsqrth xxsqrt2color(white)("dd") =color(white)("dd") hsqrt(hxx2)color(white)("dd") =color(white)("dd") hsqrt(2h))#
#color(white)("d")#
#color(brown)("and "2sqrt(2) xxsqrt2color(white)("d") =color(white)("d") 2sqrt(2xx2)color(white)("d") =color(white)("d") 2sqrt(4)color(white)("d") =color(white)("d") 2xx2=4#
Technically the answer should be #+-(hsqrt(2h))/4#
As the square root of a number is #+-#
Example:
#(-2)xx(-2)color(white)("d")=color(white)("d")(+4)color(white)("d")=color(white)("d")(+2)xx(+2)#
