show that the roots of equation (a + b)x2 -2(a2-b2)x + (a-b)2=0 are equal?

For the original question :
(a + b)x² - 2(a² - b²)x + (a - b)²=0

Discriminant of the equation, Δ
= [-2(a² - b²)]² - 4(a + b)(a - b)²
= (-2)²(a² - b²)² - 4(a + b)(a - b)²
= 4[(a + b)(a - b)]² - 4(a + b)(a - b)²
= 4(a + b)²(a - b)² - 4(a + b)(a - b)²
= 4(a + b)(a - b)²[(a + b) - 1]
= 4(a + b)(a - b)²(a + b - 1)
≠ 0

Since discriminant ≠ 0, then the roots of the equation are NOT equal.


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If the equation is (a + b)²x² - 2(a² - b²)x + (a - b)²=0 instead

Discriminant of the equation, Δ
= [-2(a² - b²)]² - 4(a + b)²(a - b)²
= (-2)²(a² - b²)² - 4(a + b)²(a - b)²
= 4[(a + b)(a - b)]² - 4(a + b)²(a - b)²
= 4(a + b)²(a - b)² - 4(a + b)²(a - b)²
= 0

Since discriminant = 0, then the roots of the equation are equal.

False discriminant =/= 0

... are NOT equal!
https://www.wolframalpha.com/input/?i=discriminant%5B(a%2Bb)*x%5E2-2*(a%5E2-b%5E2)*x%2B(a-b)%5E2,x%5D

(a+b)x^2 - 2(a^2-b^2)x + (a-b)^2 = 0...[1]. Roots are equal iff discriminant = 0.
Discriminant = d, say, = 4[(a^2-b^2)^2 - (a+b)(a-b)^2]. Then (d/4) = [(a-b)^2{(a+b)^2 -(a+b)}]
= (a+b)(a+b-1)(a-b)^2. Clearly, d is not identically = 0. Therefore roots of [1] are NOT equal.