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Let a^2=\frac{16}{44}$ and $b^2=\frac{(2+\sqrt{5})^2}{11}$, where $a$ is a negative real number and $b$ is a positive real number. If (a+b)^3 can be expressed in the simplified form $\frac{x\sqrt{y}}{z}$ where $x$, $y$, and $z$ are positive integers, what is the value of the sum $x+y+z$?

 May 22, 2019

Best Answer 

 #1
avatar+8853 
+3

Let  \(a^2=\frac{16}{44}\)  and  \(b^2=\frac{(2+\sqrt{5})^2}{11}\) , where  \(a\)  is a negative real number and  \(b\)  is a positive real number.

If  \((a+b)^3\)  can be expressed in the simplified form  \(\frac{x\sqrt{y}}{z}\)  where  \(x\),  \(y\),  and  \(z\)  are positive integers,

what is the value of the sum  \(\vphantom{\frac{x\sqrt{y}}{z}}x+y+z\) ?

______________________________________

 

\(a\quad=\quad-\sqrt{\frac{16}{44}}\quad=\quad -\frac{4}{2\sqrt{11}}\quad=\quad\frac{-2}{\sqrt{11}}\\~\\ b\quad=\quad\sqrt{\frac{(2+\sqrt{5})^2}{11}}\quad=\quad\frac{2+\sqrt{5}}{\sqrt{11}}\\~\\~\\ (a+b)^3\,=\,\Big(\frac{-2}{\sqrt{11}}+\frac{2+\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\,\Big(\frac{-2+2+\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\,\Big(\frac{\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\, \frac{\sqrt{5}}{\sqrt{11}} \cdot\frac{\sqrt{5}}{\sqrt{11}}\cdot\frac{\sqrt{5}}{\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{5}}{11\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{5}}{11\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{55}}{121}\)

 

Now it is in the form  \( \frac{x\sqrt{y}}{z}\)  where  x,  y,  and  z  are positive integers.

 

x + y + z  =  5 + 55 + 121  =  181

 May 22, 2019
 #1
avatar+8853 
+3
Best Answer

Let  \(a^2=\frac{16}{44}\)  and  \(b^2=\frac{(2+\sqrt{5})^2}{11}\) , where  \(a\)  is a negative real number and  \(b\)  is a positive real number.

If  \((a+b)^3\)  can be expressed in the simplified form  \(\frac{x\sqrt{y}}{z}\)  where  \(x\),  \(y\),  and  \(z\)  are positive integers,

what is the value of the sum  \(\vphantom{\frac{x\sqrt{y}}{z}}x+y+z\) ?

______________________________________

 

\(a\quad=\quad-\sqrt{\frac{16}{44}}\quad=\quad -\frac{4}{2\sqrt{11}}\quad=\quad\frac{-2}{\sqrt{11}}\\~\\ b\quad=\quad\sqrt{\frac{(2+\sqrt{5})^2}{11}}\quad=\quad\frac{2+\sqrt{5}}{\sqrt{11}}\\~\\~\\ (a+b)^3\,=\,\Big(\frac{-2}{\sqrt{11}}+\frac{2+\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\,\Big(\frac{-2+2+\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\,\Big(\frac{\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\, \frac{\sqrt{5}}{\sqrt{11}} \cdot\frac{\sqrt{5}}{\sqrt{11}}\cdot\frac{\sqrt{5}}{\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{5}}{11\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{5}}{11\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{55}}{121}\)

 

Now it is in the form  \( \frac{x\sqrt{y}}{z}\)  where  x,  y,  and  z  are positive integers.

 

x + y + z  =  5 + 55 + 121  =  181

hectictar May 22, 2019

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