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.  

Calculate the determinant:$$4(a+b)^2-4(a-b)^2 = 16ab$$
Then find the roots as following:
$$x^2 = \frac{2(a+b)+\sqrt{16ab}}{2},\frac{2(a+b)-\sqrt{16ab}}{2}$$
$$x^2 = a+b+2\sqrt {ab}, a+b-2\sqrt {ab}$$
$$x = \pm(\sqrt a + \sqrt b), \pm(\sqrt a - \sqrt b)$$
