x's and y's

Algebra Level 2

If x x and y y are real numbers such that x 3 + 3 y 3 x 3 − 3 y 3 = 212 131 \dfrac{x^3+3y^3}{x^3-3y^3} = \dfrac{212}{131} , what is the value of x + 2 y x − 2 y \dfrac{x+2y}{x-2y} ?


The answer is 13.

Anuj Shikarkhane by Anuj Shikarkhane , 19, 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

Chew-Seong Cheong
Jun 13, 2018

x 3 + 3 y 3 x 3 − 3 y 3 = 212 131 Given 131 x 3 + 393 y 3 = 212 x 3 − 636 y 3 81 x 3 = 1029 y 3 Divide both sides by 3 27 x 3 343 y 3 Take cube root on both sides 3 x = 7 y \begin{aligned} \frac {x^3+3y^3}{x^3-3y^3} & = \frac {212}{131} & \small \color{#3D99F6} \text{Given} \\ 131x^3 + 393y^3 & = 212x^3-636y^3 \\ 81x^3 & = 1029y^3 & \small \color{#3D99F6} \text{Divide both sides by }3 \\ 27x^3 & 343y^3 & \small \color{#3D99F6} \text{Take cube root on both sides} \\ 3x & = 7y \end{aligned}

Now, we need to find:

X = x + 2 y x − 2 y Multiply up and down by 3 = 3 x + 6 y 3 x − 6 y Note that 3 x = 7 y = 7 y + 6 y 7 y − 6 y = 13 y y = 13 \begin{aligned} X & = \frac {x+2y}{x-2y} & \small \color{#3D99F6} \text{Multiply up and down by }3 \\ & = \frac {3x+6y}{3x-6y} & \small \color{#3D99F6} \text{Note that }3x = 7y \\ & = \frac {7y+6y}{7y-6y} \\ & = \frac {13y}{y} \\ & = \boxed{13} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Will it be easy if we use componendo and dividendo?

A Former Brilliant Member - 2 years, 10 months ago

Log in to reply

Yes it will be easy.

Akshaj Garg - 1 year, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...