I assume the equation is
√(2x - 5) - √(x + 6) = 0
Note the domains for the root expressions:
• √(2x - 5) : 2x - 5 ≥ 0 ⇒ x ≥ 5/2
• √(x + 6) : x + 6 ≥ 0 ⇒ x ≥ -6
So any valid solution we find must be at least 5/2.
Move one term to the other side.
√(2x - 5) = √(x + 6)
Take squares.
(√(2x - 5))² = (√(x + 6))²
2x - 5 = x + 6
Solve for x :
x = 11
I^3=-i
therefor -27i=(3i)^3
so we can get a sum of 2 perfect cubes
a^3+b^3=(a+b)(a^2-ab+b^2)
so
remember that i²=-1
x^3+(3i)^3=(x+3i)(x²-3xi-1)
1,6,7 are positive
2,9 are negative
3,4,5,8 are undefined
