b^3-3b=0

b(b^2-3)=0

b(b-\sqrt{3})(b+\sqrt{3})=0

b \in (0,\sqrt{3},-\sqrt{3})

frac{\partial (\rho e)}{\partial t} + (\rho e+p)\frac{\partial u_{i}}{\partial x_{i}} = \frac{\partial(\tau_{ij}u_{j})}{\partial x_{i}} + \rho f_{i}u_{i} + \frac{\partial(\dot{ q_{i}})}{\partial x_{i}} + r

\sqrt{3}

-\sqrt{3}

a = 1
b = 27
while True:
	if abs(a*3 - a*a*a) > abs(b*3 - b*b*b):
		a = (a + b) / 2
	else:
		b = (a + b) / 2

	print(a,"  ",b)

a = 1
b = 27

while True:
    diff = abs(a*3 - a*a*a) - abs(b*3 - b*b*b)
    if abs(diff) < 1e-6:
        break
    
    if diff > 0: 
        a = (a + b) / 2
    else:
        b = (a + b) / 2

print(a)

b + b + b = b \cdot b  \cdot b \\ 
  3b =  b^3 \\
  3b - b^3 = 0 \\
  b(3-b^2) = 0 \\
  b_0=0 \,,\, b_1 =\sqrt{3} \,,\, b_2=-\sqrt{3}

4b = b^4

x=x

from itertools import combinations
print (list(combinations([1, 2, 3, 4], 2)))
#> [(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

\LaTeX

\prod_{i=14}^{88} n_i^2

\Large \text{Is it possible for an irrational number raised to the power of an irrational number to be a rational number?}

x + x + x = x * x * x
3x = x**3
3 = x**3 / x
3 = x**2
sqrt(3) = x

\sqrt[\pi]{1000}

\pi