Getting rid of the root $(x-\sqrt[3]{x^3-1})$

Question

How should I get rid of the 3rd root $$(x-\sqrt[3]{x^3-1})?$$

I know how to use $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$ but what can I do in this case?

score 3 · Accepted Answer · answered Mar 01 '17 at 16:56

3

HINT:

Recall that

$$x^3-y^3=(x-y)(x^2+xy+y^2)$$

Now, let $y=\sqrt[3]{x^3-1}$

answered Mar 01 '17 at 16:56

Mark Viola

But will this "get rid of the cube root"? Granted, the OP is not very clear about what they want. Your suggestion works fine if they, for example, want to compute the limit of this as $x\to\infty$ (and possibly for other things), but then the question would become a near duplicate of oldies such as this. – Jyrki Lahtonen Mar 01 '17 at 17:06
@JyrkiLahtonen Yes you are right, I meant to make it easier to calculate the limit – gbox Mar 01 '17 at 17:07
Ahh, ok. Thanks for the clarification @gbox! The technique is also used here and in an ultimate form here. – Jyrki Lahtonen Mar 01 '17 at 17:09
@JyrkiLahtonen Yes, of course it doesn't "get rid of the cube root," it merely rearranges things. However, reading the OP, it states explicitly "I know how to use $(\sqrt a +\sqrt b))(\sqrt a - \sqrt b)=a-b$, but what can I do in this case?" So, it seemed evident that the OP was seeking a way to factor. Does that make sense? -Mark – Mark Viola Mar 01 '17 at 17:13
Yup. All clear now. – Jyrki Lahtonen Mar 01 '17 at 17:15

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