Question

Hey guys- need help in this one

Answer 1

Answer:

c

Step-by-step explanation:

$x=5+i\sqrt{7} , y = 5-i\sqrt{7}\\ x^2-2xy+y^2=(x-y)^2\\ (x-y)^2=(2i\sqrt{7} )^2\\4.i^2.7\\-28$

Answer 2

Answer:

c

Step-by-step explanation:

given x = 5 + i$\sqrt{7}$ then y = 5 - i$\sqrt{7}$

x² - 2xy + y² ← is a perfect square

= (x - y)² ← substitute values for x and y

= (5 + i$\sqrt{7}$ - (5 - i$\sqrt{7}$ )²

=  (5 + i$\sqrt{7}$ - 5 + i$\sqrt{7}$ )²

= (2i$\sqrt{7}$ )²

= 2² × i² × ($\sqrt{7}$ )²

= 4 × - 1 × 7

= - 28