#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

Add 8 to both sides:

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

Add 1 to both sides:

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

#1**+10 **

Best Answer

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

Add 8 to both sides:

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

Add 1 to both sides:

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

#3**+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

CPhill Jul 14, 2016