<s>[INST] Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ Proving the roots of 

$$x^4-2(a+b)x^2+(a-b)^2=0$$
  are...... 
  $$x=\sqrt{a}+\sqrt{b}$$
  $$x=\sqrt{a}-\sqrt{b}$$
  $$x=-\sqrt{a}+\sqrt{b}$$
  $$x=-\sqrt{a}-\sqrt{b}$$
  When $a$ and $b$ are  real numbers(negative or positive) 

I  proved this by substituting $x=\sqrt{a}+\sqrt{b}$ in the main equation , but I think this way not effective because  It needs to be repeated four times to complete the proving.
Is there another effective proving.  
 [/INST] One way to verify this with minimal computations is to expand the product $P(x) = \prod (x\pm\sqrt{a}\pm\sqrt{b})$ (where the product runs over all combinations of signs).
Taking as a variable $t_1 = x+\sqrt{a}$ and $t_2 = x-\sqrt{a}$, this becomes
$$P(x) = (t_1 + \sqrt{b})(t_1-\sqrt{b})(t_2 + \sqrt{b})(t_2-\sqrt{b}) = (t_1^2 - b)(t_2^2 - b).$$
Now $t_1t_2 = x^2-a$ and $t_1^2 + t_2^2 = 2(x^2 + a)$. So 
$$P(x) = (x^2-a)^2 - 2b(x^2+a) + b^2 = x^4 - 2(a+b)x^2 +a^2 - 2ab + b^2. $$
This being said, the from math110's answer is nicer. In fact if you read it like this, it is even more obvious.
$$ \begin{aligned} x = \sqrt{a} + \sqrt{b} & \overset{\mathrm{square}}\implies x^2 = a^2 + b^2 + 2\sqrt{ab} \\ &\overset{\mathrm{rewrite}}\implies x^2 - a^2 - b^2 = 2\sqrt{ab} \\ &\overset{\mathrm{square}}\implies (x^2-a^2-b^2)^2 = 4ab. \end{aligned}$$
</s>