+0  
 
0
399
5
avatar

\( Let $x$ and $y$ be real numbers whose absolute values are different and that satisfy \begin{align*} x^3 &= 20x + 7y \\ y^3 &= 7x + 20y \end{align*} Find $xy.$\)\( \[\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\]\)1.

 

\( $a^3 + \dfrac{1}{a^3}$ if $a+\dfrac{1}{a} = 6$\)2.

 

 

 

ANNNNNNDDDDDD!

\( \begin{align*} x^3 &= 20x + 7y \\ y^3 &= 7x + 20y \end{align*}\)Find $xy$

Guest Jun 8, 2017
 #1
avatar
0

Ignore that text at the top :)

Guest Jun 8, 2017
 #4
avatar+473 
0

But why did you even put it?

AsadRehman  Jun 9, 2017
 #2
avatar+89972 
+1

 

 

a^3  +  1/a^3    if    a + 1/a  = 6

 

Note:  if  a + 1/a   = 6   then

 

(a + 1/a)^2  = 36

a^2 + 2 + 1/a^2  = 36

a^2 + 1/a^2  =  34

 

Factor  a^3  +  1/a^3   as

 

(a +  1/a) ( a^2 - 1 + 1/a^2 )  =

 

( a + 1/a) ( [a^2 + 1/a^2] - 1)  =

(6) ( [34] - 1)  =

(6) (33)  =

198

 

 

cool cool cool

CPhill  Jun 8, 2017
 #3
avatar+89972 
+1

x^3  = 20x + 7y   

y^3  = 7x + 20y

 

Subtract the two equations

x^3  - y^3  = 13x - 13y

(x-y) (x^2 + xy + y^2)  = 13 (x - y)    divide both sides by ( x - y)

x^2 +  xy + y ^2   =  13  →  x^2 + y^2  =  13 - xy   (1)

 

Add the two equations

x^3 + y^3  =  27x + 27y

(x + y) ( x^2  - xy + y^2)  = 27 (x + y)   divide both sides by (x + y)

x^2 - xy + y^2  =  27  →  x^2 + y^2  =  27 + xy    (2)

 

Then.....setting (1)  and (2)   equal, we have that

 

13 - xy  =  27 + xy

 

2xy  = -14

 

xy  =  -7

 

[ Note....other solutions  are possible...for instance  the trivial solution of (x, y) = (0, 0)  produces xy = 0 ]

 

 

 

cool cool cool

CPhill  Jun 8, 2017
 #5
avatar+7024 
+1

\(\because x^3=20x+7y\text{ and }y^3=7x+20y\\ \therefore x^3-y^3=(x-y)(x^2+xy+y^2)=13x-13y\\ x^2+xy+y^2 = 13\\ x^3+y^3 =(x+y)(x^2-xy+y^2)=27(x+y)\\ x^2-xy+y^2=27\\ (x^2+xy+y^2)-(x^2-xy+y^2)=-14\\ xy = -7\)

MaxWong  Jun 9, 2017
edited by MaxWong  Jun 9, 2017

25 Online Users

avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.