Let’s resolve into factors

x²-2ax + (a + b)(a-b)

x²-2ax + (a + b)(a-b) … (1)

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q =-2a … (3)

pq =(a + b)(a-b) … (4)

to make the factors of 3 from 4

-2a=-(a-b)-(a + b)

∴ (1) can be written as

x²-(a-b)x-(a + b)x + (a + b)(a-b)

=[{x-(a-b)}{x-(a + b)}]

∴ the resolved factors are [{x-(a-b)}{x-(a + b)}]