I'm stuck...help

Solve for a:
8 a^6-19 a^3-27 = 0

 

Substitute x = a^3:
8 x^2-19 x-27 = 0

 

The left hand side factors into a product with two terms:
(x+1) (8 x-27) = 0

 

Split into two equations:
x+1 = 0 or 8 x-27 = 0

 

Subtract 1 from both sides:
x = -1 or 8 x-27 = 0

 

Substitute back for x = a^3:
a^3 = -1 or 8 x-27 = 0

 

Taking cube roots gives (-1)^(1/3) times the third roots of unity:
a = -1 or a = (-1)^(1/3) or a = -(-1)^(2/3) or 8 x-27 = 0

 

Add 27 to both sides:
a = -1 or a = (-1)^(1/3) or a = -(-1)^(2/3) or 8 x = 27

 

Divide both sides by 8:
a = -1 or a = (-1)^(1/3) or a = -(-1)^(2/3) or x = 27/8

 

Substitute back for x = a^3:
a = -1 or a = (-1)^(1/3) or a = -(-1)^(2/3) or a^3 = 27/8

 

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

 

Because your equation is of 6th degree, you will have 6 differrent answers.

\(a^3=x\\ 8x^2-19x-27=0\\ (x+1)(8x-27)\\ x_1=-1\\ x_2=\frac{27}8\\ a_1=\sqrt[3]{-1}=-1 \\ a_2=\sqrt[3]{\frac{27}{8}}=\frac32\)

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