1. Rationalizing Trinomial Surds

How do I rationalize the follwoing trinomial surd ?

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

2. 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 $\displaystyle \sqrt[3]{2}$.

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

4. 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?

,

,

,

,

,

,

,

,

,

,

,

,

,

,

rationalising a trinomial surd denominator

Click on a term to search for related topics.