How to solve x³=y-2³ , y=√81

y = √81
y = 9
substitute y = 9 into
x^3 = y – 2^3
x^3 = 9 – 8
x^3 = 1
x = 3√1
x = 1
Therefore y = 9 and x = 1

x³=y-2³
y=√81................thus y=9

substitute into the first equation

x^3=9-2^3

x^3=9-8

x^3=1...........NOW this is a degree 3 equation so x will have 3 answer

x^3-1=0

FACTOR

(x-1)(x^2+x+1)=0

x-1=0.............x^2+x+1=0

x=1................solve with quadratic formula

x^2+x+1=0

x=[-(1)+-sqrt(1^2-4*1*1)]/[2(1)]

x=[-1+-sqrt(1-4)]/(2)

x=(-1+-sqrt(-3))/2

x=(-1+-isqrt3)/2

x={1,(-1+-isqrt3)/2

Sqr(81) = 9 = y.
Y = 9 & x= 1.
X3 = 9 - 8
x3 = 1
therefore cbrt(1)
= 1..

y=9 so x^3=9-8=1 giving x=cube root of 1,=1
Solution x=1,y=9

x=cis(2kpi/3) where k is a natural number