# What is x y in system sq root(x+y) + (x-y) cube root =6 6root((x+y)^3(x-y)^2)=8 (x+y)^3(x-y)^2 are unnder 6root

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### 1 Answer

The formulation of problem is equivocal because of misuse of mathematical symbols, hence, supposing that you want to solve the following system of equations, yields:

`{(sqrt(x+y) + root(3)(x - y) = 6),(root(6)((x+y)^3(x-y)^2) = 8):}`

You need to re-write the bottom equation such that:

`root(6)((x+y)^3(x-y)^2) = (x+y)^(3/6)*(x-y)^(2/6)`

`root(6)((x+y)^3(x-y)^2) = (x+y)^(1/2)*(x-y)^(1/3)`

`root(6)((x+y)^3(x-y)^2) = sqrt(x + y)*root(3)(x - y)`

You should come up with the following substitution, such that:

`{(sqrt(x + y) = t), (root(3)(x - y) = v):}`

Changing the variables, yields:

`{(t + v = 6),(tv = 8):}`

You need to replace `6 - t` for `v` in the bottom equation, such that:

`{(v = 6 - t),(t(6 - t) = 8):} => {(v = 6 - t),(-t^2 + 6t = 8):} => {(v = 6 - t),(t^2 - 6t + 8 = 0):}`

You should use quadratic formula to evaluate `t_(1,2)` , such that:

`t_(1,2) = (6+-sqrt(36 - 32))/2 => t_(1,2) = (6+-2)/2`

`t_1 = 4 ; t_2 = 2 `

`v_1 = 2; v_2 = 4`

Replacing 4 for `sqrt(x + y)` and 2 for `root(3)(x - y)` yields:

`sqrt(x + y) = 4 => x + y = 16`

`root(3)(x - y)= 2 => x - y = 8`

Adding the equations yields:

`x + y + x - y = 24 => 2x = 24 => x = 12 => y = 4`

Replacing 2 for `sqrt(x + y)` and 4 for `root(3)(x - y)` yields:

`sqrt(x + y) = 2 => x + y = 4`

`root(3)(x - y)= 4 => x - y = 64`

Adding the equations yields:

`x + y + x - y = 68 => 2x = 68 => x = 34 => y = -30`

**Hence, evaluating the solutions to the system of equations, yields `x = 12, y = 4` and `x = 34, y = -30` .**