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# Evaluate \$a^3 + \dfrac{1}{a^3}\$ if \$a+\dfrac{1}{a} = 6\$.

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Evaluate \$a^3 + \dfrac{1}{a^3}\$ if \$a+\dfrac{1}{a} = 6\$.

Guest Sep 28, 2017

#3
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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

CPhill  Sep 29, 2017
edited by CPhill  Sep 29, 2017
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#1
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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

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)

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)

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)

Guest Sep 29, 2017
#2
0

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

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)

a = 3 + 2 sqrt(2) or a - 3 = -2 sqrt(2)

a = 3 + 2 sqrt(2)      or         a = 3 - 2 sqrt(2)

[3 +2 sqrt(2)]^3 + 1/ [3 + 2 sqrt(2)]^3 =198

Guest Sep 29, 2017
#3
+79786
+2

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

CPhill  Sep 29, 2017
edited by CPhill  Sep 29, 2017

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