# (u^3+v^3+2)+(3uv-4)(u+v)=0 ?

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### 2 Answers

You should open the brackets such that:

`u^3 + v^3 + 2 + 3uv(u+v) - 4(u+v) = 0`

You need to group the terms such that:

`(u^3 + v^3 + 3uv(u+v))- 4(u+v) = 0`

Notice that the first group represents th expansion of the following:

`(u+v)^3 = u^3 + v^3 + 3uv(u+v)`

Hence, you should substitute `(u+v)^3` for `u^3 + v^3 + 3uv(u+v)` such that:

`(u+v)^3 - 4(u+v) = 0`

You need to factor out `u+v` such that:

`(u+v)((u+v)^2 - 4) = 0`

Setting each factor equal to zero yields:

`u + v = 0 => u = -v`

`(u+v)^2 - 4 = 0 => u+v = +-2`

**Since `u=-v` , then the second equation contradicts the first, hence, the solutions to the given equation are `u,v in R` .**

(u^3+v^3+2)+(3uv-4)(u+v)=0 ???

u^3+v^3+2+3u^v+3uv^2-4v-4u

(a + *b*)*3* = a3 + 3a2b + 3ab2 + *b3*

*so(u+v)^3-4(u+v)+2*

*(u+v)^3-2(2(u+v)+1)*