Math Q (1 Viewer)

axwe7 · Dec 23, 2015

Thanks.

InteGrand · Dec 23, 2015

$\noindent (a) We have$

$\begin{align*}\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&= \frac{1}{\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}} \\ &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left[ \left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5} \right] \left[ \left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5} \right] } \quad (\text{multiplying top and bottom by }\sqrt{2}+\sqrt{3}-\sqrt{5}) \\ &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}\right)^2 - 5} \quad (\text{difference of two squares in the denominator}) \\ &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+2\sqrt{6}+3-5} \quad (\text{expanding denominator}) \\ &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}\\ &= \frac{\sqrt{6} \left(\sqrt{2}+\sqrt{3}-\sqrt{5} \right)}{12}.\end{align*}$

$\noindent (b) We have$

$\begin{align*}\frac{1}{\sqrt[3] {2}} &= \frac{1}{\sqrt[3]{2}}\times \frac{\sqrt[3]{2}^2}{\sqrt[3]{2}^2} \\ &= \frac{\sqrt[3]{2}^2}{2}. \end{align*}$

$\noindent (c) We know that $a^3 -1 = (a-1)\left(a^2 + a +1\right)$. Let $a=\sqrt[3] {2}$, so $a^3 = 2$, then$

$\begin{align*}\frac{1}{\sqrt[3]{2} -1}&= \frac{1}{a-1}\times \frac{a^2+a+1}{a^2 + a + 1} \\ &= \frac{a^2 + a + 1}{a^3 -1 }\\ &= \frac{\sqrt[3]{2}^2 + \sqrt[3] {2} + 1}{2-1} \\\ &=\sqrt[3]{2}^2 + \sqrt[3] {2} + 1.\end{align*}$

KingOfActing · Dec 23, 2015

InteGrand how do you type so quickly in Latex? I must know!

InteGrand · Dec 23, 2015

KingOfActing said:
InteGrand how do you type so quickly in Latex? I must know!

I guess I just got used to it haha.

KingOfActing · Dec 23, 2015

InteGrand said:
I guess I just got used to it haha.

Seriously, I was still typing up the first part and you did all three already! Awesome stuff

leehuan · Dec 23, 2015

KingOfActing said:
Seriously, I was still typing up the first part and you did all three already! Awesome stuff

Meanwhile I still use Daum.

axwe7 · Dec 23, 2015

Thanks to everyone for the help.

Math Q (1 Viewer)

axwe7

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InteGrand

Well-Known Member

KingOfActing

lukewarm mess

InteGrand

Well-Known Member

KingOfActing

lukewarm mess

leehuan

Well-Known Member

axwe7

Member

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