An alternative is to solve for 'a' using the quadratic formula (after you get everything to one side and have ). That leads to two complex solutions in the form . After cubing both sides of the two equations you'll lead to . This method is a lot longer than the first method so I find it ideal to follow the first method.
if a^2 + 1 = a find the value of a^3.
a² + 1 = a
a² - a + 1 = 0
We recognize the left side as one of the factors in
the factorization of the sum of two cubes
a³ + 1 = (a + 1)(a² - a + 1)
So we multiply both sides by (a + 1)
(a + 1)(a² - a + 1) = 0(a + 1)
a³ + 1 = 0
a³ = -1
Edwin
What is interesting in this problem (and why I am writing these lines)
is that = -1 , BUT a =/= -1 (!!).
The value of "a" in this problem is one of the two complex cubic roots of -1.