Evaluate a^3 + 1/a^3 if a + 1/a = 6
If a + 1/a = 6 square both sides
(a^2 + 2 + 1/a^2) = 36
a^2 + 1/a^2 = 34
Note that a^3 + 1/a^3 can be factored as a sum of cubes....so we have
a^3 + 1/a^3 =
( a + 1/a) ( a^2 - 1 + 1/a^2)
(6) ( a^2 + 1/a^2 - 1 )
(6) ( 34 - 1) =
6 * 33 =
198
Solve for a:
a^3 + 1/a^3 = 6
Bring a^3 + 1/a^3 together using the common denominator a^3:
(a^6 + 1)/a^3 = 6
Multiply both sides by a^3:
a^6 + 1 = 6 a^3
Subtract 6 a^3 from both sides:
a^6 - 6 a^3 + 1 = 0
Substitute x = a^3:
x^2 - 6 x + 1 = 0
Subtract 1 from both sides:
x^2 - 6 x = -1
Add 9 to both sides:
x^2 - 6 x + 9 = 8
Write the left hand side as a square:
(x - 3)^2 = 8
Take the square root of both sides:
x - 3 = 2 sqrt(2) or x - 3 = -2 sqrt(2)
Add 3 to both sides:
x = 3 + 2 sqrt(2) or x - 3 = -2 sqrt(2)
Substitute back for x = a^3:
a^3 = 3 + 2 sqrt(2) or x - 3 = -2 sqrt(2)
Taking cube roots gives (3 + 2 sqrt(2))^(1/3) times the third roots of unity:
a = -(-3 - 2 sqrt(2))^(1/3) or a = (3 + 2 sqrt(2))^(1/3) or a = (-1)^(2/3) (3 + 2 sqrt(2))^(1/3) or x - 3 = -2 sqrt(2)
Add 3 to both sides:
a = -(-3 - 2 sqrt(2))^(1/3) or a = (3 + 2 sqrt(2))^(1/3) or a = (-1)^(2/3) (3 + 2 sqrt(2))^(1/3) or x = 3 - 2 sqrt(2)
Substitute back for x = a^3:
a = -(-3 - 2 sqrt(2))^(1/3) or a = (3 + 2 sqrt(2))^(1/3) or a = (-1)^(2/3) (3 + 2 sqrt(2))^(1/3) or a^3 = 3 - 2 sqrt(2)
Taking cube roots gives (3 - 2 sqrt(2))^(1/3) times the third roots of unity:
a = -(-3 - 2 sqrt(2))^(1/3) or a = (3 + 2 sqrt(2))^(1/3) or a = (-1)^(2/3) (3 + 2 sqrt(2))^(1/3) or a = (3 - 2 sqrt(2))^(1/3) or a = (-1)^(2/3) (3 - 2 sqrt(2))^(1/3) or a = -(2 sqrt(2) - 3)^(1/3)
Solve for a:
a + 1/a = 6
Bring a + 1/a together using the common denominator a:
(a^2 + 1)/a = 6
Multiply both sides by a:
a^2 + 1 = 6 a
Subtract 6 a from both sides:
a^2 - 6 a + 1 = 0
Subtract 1 from both sides:
a^2 - 6 a = -1
Add 9 to both sides:
a^2 - 6 a + 9 = 8
Write the left hand side as a square:
(a - 3)^2 = 8
Take the square root of both sides:
a - 3 = 2 sqrt(2) or a - 3 = -2 sqrt(2)
Add 3 to both sides:
a = 3 + 2 sqrt(2) or a - 3 = -2 sqrt(2)
Add 3 to both sides:
a = 3 + 2 sqrt(2) or a = 3 - 2 sqrt(2)
[3 +2 sqrt(2)]^3 + 1/ [3 + 2 sqrt(2)]^3 =198
Evaluate a^3 + 1/a^3 if a + 1/a = 6
If a + 1/a = 6 square both sides
(a^2 + 2 + 1/a^2) = 36
a^2 + 1/a^2 = 34
Note that a^3 + 1/a^3 can be factored as a sum of cubes....so we have
a^3 + 1/a^3 =
( a + 1/a) ( a^2 - 1 + 1/a^2)
(6) ( a^2 + 1/a^2 - 1 )
(6) ( 34 - 1) =
6 * 33 =
198