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If (x+1/x)^2=3, what is x^3+1/x^3

Guest May 15, 2017
 #1
avatar+87301 
+1

(x+1/x)^2 = 3      expand the left side

 

x^2 + 2 +  1/x^2  = 3         subtract 3 from both sides

 

x^2 -  1  +  1/x^2   =  0         (1)

 

Now      x^3   +  1/x^3    is   a sum of cubes  which can be factored as

 

(x +  1/x)  ( x^2   -  1  +  1/x^2 )         sub (1)  into this factorization 

 

( x + 1/x)   (  0 )     =

 

0

 

 

cool cool cool

CPhill  May 15, 2017
 #2
avatar+19639 
+1

If (x+1/x)^2=3, what is x^3+1/x^3

 

\(\begin{array}{|rcll|} \hline \left(x+\frac{1}{x}\right)^2 &=&3 \\ x+\frac{1}{x} &=& \sqrt{3} \\ \hline \end{array} \)

 

\(\small{ \begin{array}{|rclcll|} \hline (x+\frac{1}{x})^3 &=& \left(x+\frac{1}{x}\right)^2 \cdot \left(x+\frac{1}{x}\right) &=& 3\cdot \sqrt{3} \\ x^3+3x^2\cdot \frac{1}{x} +3x\cdot \frac{1}{x^2}+\frac{1}{x^3} &=& && 3\cdot \sqrt{3} \\ x^3+3x +3\cdot \frac{1}{x}+\frac{1}{x^3} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3}+ 3x +3\cdot \frac{1}{x} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3}+ 3\cdot(x + \frac{1}{x}) &=& && 3\cdot \sqrt{3} \quad & | \quad x + \frac{1}{x}=\sqrt{3} \\ x^3+\frac{1}{x^3}+ 3\cdot\sqrt{3} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3} &=& && 3\cdot \sqrt{3}- 3\cdot\sqrt{3} \\ \mathbf{x^3+\frac{1}{x^3}} &\mathbf{=} & &&\mathbf{ 0 } \\ \hline \end{array} }\)

 

laugh

heureka  May 16, 2017

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