# solve the system (x^3)(y^3)(z^4)=1 and (x^2)(y^4)(z^4)=2 and (x^2)(y^3)(z^5)=3

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`(x^3)(y^3)(z^4)= 1................(1)`

`(x^2)(y^4)(z^4)= 2..................(2)`

`(x^2)(y^3)(z^5)= 3 ....................(3)`

First, we will divide (1) by (2).

==> `x/y = 1/2 ==> y= 2x ..................(4)`

``Now we will divide (1) by (3).

`==> x/z = 1/3 ==> z = 3x...................(5)`

Now, we will substitute (5) and (5) into (1).

`==> (x^3)(y^3)(z^4) = 1`

`==> (x^3)(2x)^3 (3x)^4 = 1`

`==> (x^3)(8x^3)(81x^4)= 1`

`==> 648x^10 = 1`

`==> x^10 = 1/648`

`==> x = (1/648)^(1/10)`

`==> y= 2x = 2/(648)^(1/10)`

`==> z= 3x = 3/ 648^(1/10)`

``

I propose the approach: divide the top equation by the middle equation.

(x^3)(y^3)(z^4)/(x^2)(y^4)(z^4) = 1/2 (reduce the powers that have the same base)

x/y=1/2 (multiply the means and the extremes) =>y=2x

Divide the top equation by theĀ bottom equation.

x/z=1/3 (multiply the means and the extremes) =>z=3x

use y=2x and z=3x in the top equation

(x^3)(8x^3)(81x^4)=648x^10=1 => x^10=1/648 =>x=1/(648)^(1/10)

Answer: x=1/(648)^(1/10); y= 2/(648)^(1/10); z= 3/(648)^(1/10)