+0

# Hexic Equation(Polynomial of degree 6)

0
614
3
+7220

\(a^6-9a^3+8=0\\ \textbf{Solve for a}\)

Jul 14, 2016
edited by MaxWong  Jul 14, 2016
edited by MaxWong  Jul 14, 2016
edited by MaxWong  Jul 14, 2016

#1
+10

Solve for a:
a^6-9 a^3+8 = 0

Substitute x = a^3:
x^2-9 x+8 = 0

The left hand side factors into a product with two terms:
(x-8) (x-1) = 0

Split into two equations:
x-8 = 0 or x-1 = 0

x = 8 or x-1 = 0

Substitute back for x = a^3:
a^3 = 8 or x-1 = 0

Taking cube roots gives 2 times the third roots of unity:
a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or x-1 = 0

a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or x = 1

Substitute back for x = a^3:
a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or a^3 = 1

Taking cube roots gives 1 times the third roots of unity:
Answer: | a = 2   or   a = -2 (-1)^(1/3)   or   a = 2 (-1)^(2/3)   or   a = 1   or   a = -(-1)^(1/3) or a = (-1)^(2/3)

Jul 14, 2016

#1
+10

Solve for a:
a^6-9 a^3+8 = 0

Substitute x = a^3:
x^2-9 x+8 = 0

The left hand side factors into a product with two terms:
(x-8) (x-1) = 0

Split into two equations:
x-8 = 0 or x-1 = 0

x = 8 or x-1 = 0

Substitute back for x = a^3:
a^3 = 8 or x-1 = 0

Taking cube roots gives 2 times the third roots of unity:
a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or x-1 = 0

a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or x = 1

Substitute back for x = a^3:
a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or a^3 = 1

Taking cube roots gives 1 times the third roots of unity:
Answer: | a = 2   or   a = -2 (-1)^(1/3)   or   a = 2 (-1)^(2/3)   or   a = 1   or   a = -(-1)^(1/3) or a = (-1)^(2/3)

Guest Jul 14, 2016
#2
+7220
+5

Thx guest :D

By the way, why don't you register to be a forum member? I see you answering everywhere in the forum.

MaxWong  Jul 14, 2016
#3
+96187
+5

a^6 - 9a^3 + 8 = 0      factoring directly, we have

(a^3 - 8) (a^3 -1)  = 0

Factoring these as differences of cubes, we have

(a - 2)(a^2 + 2a + 4)(a - 1)(a^2 + a + 1)  = 0

Setting the first and third linear factors to 0 produces two real solutions....a =  2  and a = 1

Setting the second factor to 0 and using thequadratic formula produces two complex solutions :

a =  [ -2 ± √ ( 4 - 16) / 2 =  [ -2 ± √ [( - 12)] / 2 =  [ -2 ± 2√ ( -3 ) ] / 2 =     -1 ± i√3

Setting the fourth factor to 0 and using thequadratic formula produces the two remaining complex solutions :

a =    [-1 ± √ (1 - 4) ] / 2  =  [ -1  ± √ (-3) ] / 2  =  [ -1  ± i√ (3) ] / 2

Jul 14, 2016