If you substitute any of the four answer choices into the equation:
(y-2)^3=(y+2)^3
you will find that none of them is a good answer.
So I suspect there is a typo for the question.
The equation as shown above can be solved by assuming
y=a+bi
Expand both sides to give
12b^2-24abi-12a^2-16=0
from which b can be solved by assuming a=any number, including zero.
If (y-2)hole power 3 = (y+2)hole power 3 then find the value of Y.
a) -16
b) -4
c) O
d) 4
2 answers
(y-2)^3 = (y+2)^3.
(y-2)^2(y-2) = (y+2)^2(y+2).
(y^2-4y+4)(y-2) = (y^2+4y+4)(y+2)
y^3-4y^2+4y-2y^2+8y-8=y^3+4y^2+4y+2y^2+8y+8,
-6y^2+12y-8 = 6y^2+12y+8,
-12y^2 = 16, Y = 1.155i.
No real solution, because y-2 cannot equal y+2.
(y-2)^2(y-2) = (y+2)^2(y+2).
(y^2-4y+4)(y-2) = (y^2+4y+4)(y+2)
y^3-4y^2+4y-2y^2+8y-8=y^3+4y^2+4y+2y^2+8y+8,
-6y^2+12y-8 = 6y^2+12y+8,
-12y^2 = 16, Y = 1.155i.
No real solution, because y-2 cannot equal y+2.