\( Let $x$ and $y$ be real numbers whose absolute values are different and that satisfy \begin{align*} x^3 &= 20x + 7y \\ y^3 &= 7x + 20y \end{align*} Find $xy.$\)\( \[\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\]\)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 $xy$
+129918
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
+129918 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 ]
