Cubic Conundrum

Algebra Level 5

Let x x , y y be real numbers other than 0 that satisfy x 3 + y 3 + 3 x 2 y 2 = x 3 y 3 x^3+y^3+3x^2y^2=x^3y^3 . Find the sum of all possible values of 1 x + 1 y \dfrac 1x+ \dfrac 1y .


The answer is -1.

Danish Ahmed by Danish Ahmed , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
Nov 1, 2020

x 3 + y 3 + 3 x 2 y 2 = x 3 y 3 x − 3 + y − 3 + 3 x − 1 y − 1 = 1 ( x − 1 + y − 1 ) 3 − 1 − 3 x − 1 y − 1 ( x − 1 + y − 1 − 1 ) = 0 ( x − 1 + y − 1 − 1 ) [ ( x − 1 + y − 1 ) 2 + x − 1 + y − 1 + 1 − 3 x − 1 y − 1 ] = 0 ( x − 1 + y − 1 − 1 ) ( x − 2 + y − 2 − x − 1 y − 1 + x − 1 + y − 1 + 1 ) = 0 1 2 ( x − 1 + y − 1 − 1 ) [ ( x − 1 − y − 1 ) 2 + ( x − 1 + 1 ) 2 + ( y − 1 + 1 ) 2 ] = 0 \begin{aligned} x^3 + y^3 + 3x^2y^2 & = \; x^3y^3 \\ x^{-3} + y^{-3} + 3x^{-1}y^{-1} & = \; 1 \\ \left(x^{-1} + y^{-1}\right)^3 - 1 - 3x^{-1}y^{-1}\left(x^{-1} + y^{-1} - 1\right) & = \; 0 \\ \left(x^{-1} + y^{-1} - 1\right)\left[\left(x^{-1} + y^{-1}\right)^2 + x^{-1} + y^{-1} + 1 - 3x^{-1}y^{-1}\right] & = \; 0 \\ \left(x^{-1} + y^{-1} - 1\right)\left(x^{-2} + y^{-2} - x^{-1}y^{-1} + x^{-1} + y^{-1} + 1\right) & = \; 0 \\ \tfrac12\left(x^{-1} + y^{-1} - 1\right)\left[\left(x^{-1} - y^{-1}\right)^2 + (x^{-1}+1)^2 + (y^{-1} + 1)^2\right] & = \; 0 \end{aligned} so that either x − 1 + y − 1 = 1 x^{-1} + y^{-1} = 1 or else x = y = − 1 x=y=-1 . Thus the sum of possible values of x − 1 + y − 1 x^{-1} + y^{-1} is 1 + − 2 = − 1 1 + -2 = \boxed{-1} .

1 Helpful 3 Interesting 9 Brilliant 0 Confused

For example x=2 & y=2 The value has questioned is equal 1 !!!

Masoud Golshan - 5 months, 2 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...