DanieldeLucena
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Rationalize the denominator of [attachment=0:ckg0ou25]1.JPG[/attachment:ckg0ou25]
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Denominator
- Thread starter DanieldeLucena
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DanieldeLucena said:Rationalize the denominator of [attachment=0:3ex65yw9]1.JPG[/attachment:3ex65yw9]
If you could get that to be 2[sup:3ex65yw9]3[/sup:3ex65yw9] - (2[sup:3ex65yw9]1/3[/sup:3ex65yw9])[sup:3ex65yw9]3[/sup:3ex65yw9], you'd no longer have a radical in the denominator.
I suggest that you look at the pattern for factoring a difference of two cubes.
a[sup:3ex65yw9]3[/sup:3ex65yw9] - b[sup:3ex65yw9]3[/sup:3ex65yw9] = (a - b)(a[sup:3ex65yw9]2[/sup:3ex65yw9] + ab + b[sup:3ex65yw9]2[/sup:3ex65yw9])
What you currently have in the denominator looks like the (a - b) part.... If you multiply numerator and denominator by something in the form of that TRINOMIAL factor, you should end up with a rational denominator.
DanieldeLucena
New member
- Joined
- Jul 29, 2010
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- 39
I'm having trouble to get this
2 - (2¹/³)
into this
2³ - (2¹/³)³
2 - (2¹/³)
into this
2³ - (2¹/³)³
As Mrspi said, the difference of two cubes is:
\(\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
In your problem \(\displaystyle a=2\) and \(\displaystyle b=\sqrt[3]{2}\)
In other words, just plug these numbers in to the equation above.
Does this help?
\(\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
In your problem \(\displaystyle a=2\) and \(\displaystyle b=\sqrt[3]{2}\)
In other words, just plug these numbers in to the equation above.
Does this help?