I need help solving this. thank you

1. a^4+8a^3+24a^2+32a+16

2.y^8+8y^7+28y^6+56y^5+70y^4+56y^3+28y^2+8y+1

3.x^6-3x^4y^2+3x^2y^4-y^6

hi there!

the answer is: 2 and -2

step-by-step explanation:

case 1: an absolute value of positive number will still be a positive number

so absolute value of 2 is 2

case 2: an absolute value of negative number will be positive number so absolute value of -2 is 2

a

step-by-step explanation:

given the 2 equations

- x + y = - 2 → (1)

2x - 3y = 4 → (2)

multiplying (1) by 2 and adding to (2) will eliminate the x- term

- 2x + 2y = - 4 → (3)

add (2) and (3) term by term

(2x - 2x) + (- 3y + 2y) = (4 - 4), simplifying gives

- y = 0 ⇒ y = 0

substitute y = 0 in either of the 2 equations for corresponding value of x

substituting in (2) gives

2x = 4 ( divide both sides by 2 )

x = 2

solution is (2, 0 ) → a