oh god oh god oh god no not sigfigs what did we do to deserve this. . . NO
*melts*
Oops. Did I do that?
^Tbh, I actually prefer decimals. Im a physicist so I dont mind rounding. Its only those neeky mathematicians who say pi/3 instead of 1.05
"They said this day would never come. They said our sights were set too high. They said this country was too divided, too disillusioned to ever come around a common purpose. But on this January night, at this defining moment in history, you have done what the cynics said we couldn't do." - Barack Obama.
clicky clicky clicky
Because decimals are, most of the time, approximations, and not exact values of a given situation. Decimals also introduce more room for potential errors; when exact values are mistaken, usually the answer goes from correct to really wrong really fast, and is a form of consistency for our theoretical, perfect little worlds we have in our heads. Also, we hate significant figures. To the death.
ugh...when I enter a science classroom it's just 2+2=5 for me all of the sudden ;_;
You should do your own homework. But I will do it for you this once. The easiest method is to use DeMoivre's Theorem, which presumably you know,
exp(iy) = (cos(y + 2(PI)n) + i sin(y + 2(PI)n)). n is an integer.
put x^3 = a*exp(iy). So y = 0 and a = 192/81.
Then
x = (a^(1/3))*exp(i(y/3))
x = (a^(1/3))*(cos(y/3 + 2(PI)n/3) + i sin(y/3 + 2(PI)n/3))
Here we have
y = 0
so
x = (a^(1/3))*(cos(2(PI)n/3) + i sin(2(PI)n/3))
Then stick in integers for n,
n = 0
x = (a^(1/3))*(cos(0) + i sin(0))
x = (a^(1/3)) = 4/3
n = 1
x = (a^(1/3))*(cos(2(PI)/3) + i sin(2(PI)/3))
x = (2/3)*(1 + i Sqrt(3))
n = 2
x = (a^(1/3))*(cos(4(PI)/3) + i sin(4(PI)/3))
x = (2/3)*(1 - i Sqrt(3))
n >= 3 just cycles through these solutions.
These solutions define an equilateral triangle in the complex plane. PuPu did it a really nice way, and yeah you have to get 3 answers.