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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
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Ignore that text at the top :)

Guest Jun 8, 2017
#4
+473
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But why did you even put it?

#2
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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

CPhill  Jun 8, 2017
#3
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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)

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 ]

CPhill  Jun 8, 2017
#5
+7024
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$$\because x^3=20x+7y\text{ and }y^3=7x+20y\\ \therefore x^3-y^3=(x-y)(x^2+xy+y^2)=13x-13y\\ x^2+xy+y^2 = 13\\ x^3+y^3 =(x+y)(x^2-xy+y^2)=27(x+y)\\ x^2-xy+y^2=27\\ (x^2+xy+y^2)-(x^2-xy+y^2)=-14\\ xy = -7$$

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