Question 242613
{{{root(5, 64x^8y^12)}}}<br>
To simplify this we need to find factors that are perfect powers of 5, if any.<br>
For the 64 we look for factors of 64 that are perfect powers of 5, if any. Since {{{32 = 2^5}}} and 64 = 32*2, we do have a perfect power of 5 factor in 64.<br>
For the variables we can factor out perfect powers of 5 as long as the exponent is greater than or equal to 5.<br>
{{{root(5, 64x^8y^12)}}}
{{{root(5, 32*2*x^5*x^3*y^5*y^7)}}} (There's another power of 5 in {{{y^7}}})
{{{root(5, 32*2*x^5*x^3*y^5*y^5*y^2)}}}
Now we can use the property of radicals, {{{root(a, p*q) = root(a, p)*root(a, q)}}}, to split up all these factors into separate radicals:
{{{root(5, 32)*root(5, 2)*root(5, x^5)*root(5, x^3)*root(5, y^5)*root(5, y^5)*root(5, y^2)}}}
Next, for each factor that is a perfect power of 5, we substitute in its value:
{{{2*root(5, 2)*x*root(5, x^3)*y*y*root(5, y^2)}}}
Now we use the Commutative and Associative Properties to rearrange and regroup the factors. We want the non-radical factors grouped together and the radical factors grouped together. (The convention is to put the radical at the end):
{{{(2*x*y*y)*(root(5, 2)*root(5, x^3)*root(5, y^2))}}}
which simplifies to:
{{{2xy^2\root(5, 2x^3y^2)}}}