Question

How do you rationalize the numerator?

`(root(3)(x^2)-4root(3)(x)+16)/(sqrt(x^3)+4)`

Answer 1

Multiply numerator and denominator by `(root(3)x+4)`

Explanation:

Let `root(3)x=t`, then we can write the numerator `root(3)(x^2)-4root(3)(x)+16` as `t^2-4t+16` and multiplying it by `t+4`, we get `t^3+64` using identity `(x^2-xy+y^2)(x+y)=x^3+y^3` and numerator is rationalized.

Hence `(root(3)(x^2)-4root(3)(x)+16)/(sqrt(x^3)+4)`

= `(root(3)(x^2)-4root(3)(x)+16)/(sqrt(x^3)+4)xx(root(3)x+4)/(root(3)x+4)`

= `(x+64)/((sqrt(x^3)+4)(root(3)x+4))`