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.  

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.
$$ \begin{aligned} x = \sqrt{a} + \sqrt{b} & \overset{\mathrm{square}}\implies x^2 = a^2 + b^2 + 2\sqrt{ab} \\ &\overset{\mathrm{rewrite}}\implies x^2 - a^2 - b^2 = 2\sqrt{ab} \\ &\overset{\mathrm{square}}\implies (x^2-a^2-b^2)^2 = 4ab. \end{aligned}$$
