```Question 397070
You did a pretty good job of using words to describe your expression. Using Algebra.com's formula rendering feature, the expression
{{{root(3, y^7)*root(3, 81y^8)}}}
is specified as:
root(3, y^7)*root(3, 81y^8)
surrounded by 3 left braces and 3 right braces. The "3" represents the type of root. So any kind of root can be done with this syntax.<br>
With complicated radical expressions one would often simplify the two radicals before multiplying them. But your expression is not that complicated so we will go ahead and use a property of radicals, {{{root(a, p)*root(a, q) = root(a, p*q)}}}, to multiply the two radicals. This gives us:
{{{root(3, y^7*81y^8)}}}
Multiplying the y factors together we get:
{{{root(3, 81y^15)}}}
Now we simplify the radical. Since this is a cube root we look for factors of the radicand (the expression inside a radical), that are perfect cubes. With a little effort we should find that 27 is a perfect cube, {{{3^3}}}, and it is a factor of 81. And {{{y^15}}} is a perfect cube itself since its exponent is a multiple of 3 ({{{(y^5)^3 = y^15}}}). Factoring out the perfect cubes we get:
{{{root(3, 27*3*y^15)}}}
And using the Commutative property to change the order so the perfect cubes are in front. (This is an optional but recommended step.):
{{{root(3, 27*y^15*3)}}}
Then we use the same property of radicals we used earlier. Only this time we use it in the opposite direction (right to left) to split this one radical into the product of the radicals of each of its factors:
{{{root(3, 27)*root(3, y^15)*root(3, 3)}}}
The cube roots of the perfect cubes simplify:
{{{3*y^5*root(3, 3)}}}
which is the simplified form of your original expression.<br>
Note: By using the Commutative property earlier the simplified cube roots of the perfect cubes end up in front of the remaining radical. This is the preferred way to write such an expression.```