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Well, I can't find the example on how to solve this.

If I multiply $$ \dfrac{2}{\sqrt[3]{9}+\sqrt[3]{15}+\sqrt[3]{25}} $$ with $$ \dfrac{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}}{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}} $$ or similar, it just gets even more complicated, and I get 4 terms instead of 3 in the denominator, and more can't be better...

Can someone tell me some principle by which all kinds of expressions with 3 or more terms and with different roots could be rationalised? On a test, I haven't got much time to use appropriate formula, as I can for square difference and so on, I need a principle.


marked as duplicate by Watson, Daniel W. Farlow, C. Falcon, Namaste algebra-precalculus Dec 27 '16 at 0:16

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The expression is $$ \frac{2}{\sqrt[3]{9}+\sqrt[3]{15}+\sqrt[3]{25}}$$

Note that $9=3^2,\ 15=3\cdot 5, \ 25=5^2$ so the denominator is of the form $a^2+ab+b^2$. Now you just use the formula $(a-b)(a^2+ab+b^2)=a^3-b^3$ for suitable $a,b$.

In general, when you have a $3$ term denominator, or more, if it does not have a special form (like above) it is unlikely you will be able to rationalize the fraction in one step. Note that when cubic or superior roots are involved, multiplying with an expression where you just change the sign (for example $(a+b+c)(a+b-c)$) of some term doesn't work, since squaring a cubic or superior root does not make the root dissapear.

For superior roots you may also use $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$ and for $n$ odd $a^n+b^n = (a+b)(a^{n-1}-a^{n-2}b+..-ab^{n-2}+b^{n-1})$ (in the last one the signs are alternating).

  • $\begingroup$ But what do I do when there are different kinds of roots? For example first one is 4, second 3, third 5, fourth 8th? $\endgroup$ – Guest114 Nov 11 '13 at 10:19
  • $\begingroup$ I cannot give you a general rule. If you have a combination of 3th, 4th and 5th roots I assure you that it is highly unlikely you will be able to do it by hand. $\endgroup$ – Beni Bogosel Nov 11 '13 at 10:23
  • $\begingroup$ So basically, I shouldn't expect these in a test. What about when I don't have an obvious special form? (For example they are all under the 3rd root, but not the numbers that are connected... like 3, 5 and 7.) $\endgroup$ – Guest114 Nov 11 '13 at 10:31
  • $\begingroup$ It all depends on the formulas your teacher gave you to do the rationalization. If he didn't gave you other formulas beside $(a-b)(a+b)=a^2-b^2$ and the ones for $a^3\pm b^3$ then you shouldn't worry about getting three cubic roots unrelated. If you get three square roots unrelated in the denominator then you have to use two times the formula for $a^2-b^2$. $\endgroup$ – Beni Bogosel Nov 11 '13 at 10:34

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