Rationalizing Trinomial Surds

**Arka** · November 14th 2010, 09:33 PM

How do I rationalize the follwoing trinomial surd ?

$\sqrt[3]{4} - \sqrt[3]{2} + 1$

**Prove It** · November 14th 2010, 10:58 PM

Remember that

$\displaystyle \sqrt[3]{4} = \sqrt[3]{2 \times 2} = \sqrt[3]{2^2} = (\sqrt[3]{2})^2$ .

So $\displaystyle \sqrt[3]{4} - \sqrt[3]{2} + 1 = (\sqrt[3]{2})^2 - \sqrt[3]{2} + 1$ .

This is a quadratic in $\displaystyle \sqrt[3]{2}$ .

**Arka** · November 14th 2010, 11:07 PM

But then , what will be the rationalizing factor of $\sqrt[3]{4} - \sqrt[3]{2} + 1$ ?

**HallsofIvy** · November 15th 2010, 03:02 AM

Arka, [tex]a^3- b^3= (a+ b)(a^2- ab+ b^2)[/quote]
Do you see that what you have is $a^2- ab+ b^2$ ? What are a and b?

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