Question

What is `root3 (-x^15y^9)`?

Answer 1

`root(3)(-x^15y^9) = -x^5y^3`

Explanation:

For all Real values of `a`:

`root(3)(a^3) = a`

Putting `a=-x^5y^3`, we find:

`root(3)(-x^15y^9) = root(3)((-x^5y^3)^3) = -x^5y^3`

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Footnote

It is a common error to think that a similar property holds for square roots, namely:

`sqrt(a^2) = a`

but this is only generally true when `a >= 0`.

What we can say for square roots is:

`sqrt(a^2) = abs(a)`

This works for any Real number `a`.

Real cube roots behave better in this case.

Answer 2

`root(3)(-x^15*y^9)=-x^5y^3`

Explanation:

In `root(3)(-x^15*y^9)`, we have `-1` a factor and as we are seeking cube root, let us write it as `(-1)^3`. Also, let us write `x^15=(x^5)^3` and `y^9=(y^3)^3`

Hence `root(3)(-x^15*y^9)`

= `root(3)((-1)^3*(x^5)^3*(y^3)^3)`

= `(-1)x^5y^3`

= `-x^5y^3`