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

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...

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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` .

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