Question

If `(x^2 + y^2)/(x^2 - y^2) = 17/8`, then find the value of:

1. x : y
2. `(x^3 + y^3)/(x^3 - y^3)`

Answer 1

It is given that:

`(x^2 + y^2)/(x^2 - y^2) = 17/8`

Applying componendo-dividendo.

`(x^2 + y^2 + x^2 - y^2)/(x^2 + y^2 - x^2 + y^2) = (17 + 8)/(17 - 8)`

`=> (2x^2)/(2y^2) = 25/9`

`=> x^2/y^2 = 25/9`

`=> x/y = +-5/3`

`=> x: y = 5 : 3`

2) `x/y = +- 5/3`

`x/y = 5/3`

`=> x^3/y^3 = 125/27`

Applying componendo-dividendo, we get

`(x^3 + y^3)/(x^3 - y^3) = (125 + 27)/(125 - 27)`

`=> (x^3 + y^3)/(x^3- y^3) = 152/98`

`=> (x^3 + y^3)/(x^3 - y^3) = 76/49`

or

`x/y = -5/3`

`"x"^3/"y"^3 = -125/27`

Applying componendo-dividendo, we get

`(x^3 + y^3)/(x^3 - y^3) = (-125 + 27)/(-125-27)`

`= (- 98)/-152`

`= 49/76`

Answer 2

(i) `(x^2 + y^2)/(x^2 - y^2) = 17/8`

Applying componendo-dividendo rule,

`(x^2 + y^2 + x^2 - y^2)/(x^2 + y^2 - x^2 + y^2) = (17 + 8)/(17 - 8)`

`(2x^2)/(2y^2) = (25)/(9)`

` x^2/y^2 = (25)/(9)`

`x/y = (5)/(3)`

`x: y = 5 : 3`.

(ii) `x/y =  (5)/(3)`

Taking cube on both sides,

`x^3/y^3 = (125)/(27)`

Applying componendo-dividendo rule,

`(x^3 + y^3)/(x^3 - y^3) = (125 + 27)/(125 - 27)`

`(x^3 + y^3)/(x^3- y^3) = 152/98`

`(x^3 + y^3)/(x^3- y^3) = 76/49`

`(x^3 + y^3)/(x^3- y^3) = 49/76`