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

Guest Jan 31, 2015

Best Answer 

 #3
avatar+92196 
+10

 

$$\boxed{(a^3+b^3)=(a+b)(a^2-ab+b^2)}\\\\
(a^3+b^3)=(a+b)(a^2-ab+b^2+3ab-3ab)\\\\
(a^3+b^3)=(a+b)(a^2+2ab+b^2-3ab)\\\\
(a^3+b^3)=(a+b)((a+b)^2-3ab)\\\\\\
But \;b=\frac{1}{a} \qquad and \;\;a+b=6\\\\
(a^3+\frac{1}{a}^3)=(6)((6)^2-3a*\frac{1}{a})\\\\
(a^3+\frac{1}{a}^3)=(6)(36-3)\\\\
(a^3+\frac{1}{a}^3)=(6)(33)\\\\
(a^3+\frac{1}{a}^3)=198\\\\$$

Melody  Feb 1, 2015
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5+0 Answers

 #1
avatar+85644 
+10

a + 1/a = 6   multiply through by a

a^2 + 1 = 6a    rearrange

a^2  -6a + 1 = 0    and using the quadratic formula, we find that a = 3 ±√8

 

And  a^3 + 1/a^3  factors as

(a + 1/a)(a^2 -1 + 1/a^2)      and (a + 1/a) = 6 ...so we have

6(a^2 - 1 + 1/a^2)

If a= 3 + √8 we have

6[ (3 + √8)^2 - 1 + 1 / (3 +√8)^2 ] =

6 [ (9 + 6√8 + 8 - 1 + 1/ (9 + 6√8 + 8 ] =

6 [ 16 + 6√8 + 1/ (17 + 6√8)]    and using the conjugate, (17 - 6√8), we have

6[ 16 + 6√8 + (17  - 6√8) / (289 - 288)]=

6[16 + 6√8 + (17  - 6√8) / 1] =

6[16 + 6√8 + (17  - 6√8)] =

6[ 16 + 17] = 198    

 

And if a = 3 - √8   we have

6[ (3 - √8)^2 - 1 + 1 / (3 -√8)^2 ] =

6 [ (9 - 6√8 + 8 - 1 + 1/ (9 - 6√8 + 8 ] =

6 [ 16 - 6√8 + 1/ (17 - 6√8)]    and using the conjugate, (17 + 6√8) we have

6[ 16 - 6√8 + (17  + 6√8) / (289 - 288)]=

6[16 - 6√8 + (17  + 6√8) / 1] =

6[16 - 6√8 + (17  + 6√8)] =

6[ 16 + 17] = 198 

 

Exactly the same result !!!  

And that's the evaluation of  a^3 + 1/a^3......

 

 

CPhill  Jan 31, 2015
 #2
avatar+19207 
+10

Evaluate

 a^3 + \dfrac{1}{a^3} 

if

 a+\dfrac{1}{a} = 6.

$$\left(a+\frac{1}{a}\right)^3=6^3\\\\
a^3+3a^2*\frac{1}{a}+3a*\frac{1}{a^2}+\frac{1}{a^3}=6^3\\\\
a^3+3a+3\frac{1}{a}+\frac{1}{a^3}=6^3\\\\
a^3+\frac{1}{a^3}+3a+3\frac{1}{a}=6^3\\\\
a^3+\frac{1}{a^3}+3\underbrace{\left(a+\frac{1}{a}\right)}_{=6}=6^3\\\\
a^3+\frac{1}{a^3}+3*6=6^3\\\\
a^3+\frac{1}{a^3}=6^3-3*6\\\\
a^3+\frac{1}{a^3}=6(6^2-3)\\\\
a^3+\frac{1}{a^3}=6(33)\\\\
a^3+\frac{1}{a^3}=198$$

heureka  Feb 1, 2015
 #3
avatar+92196 
+10
Best Answer

 

$$\boxed{(a^3+b^3)=(a+b)(a^2-ab+b^2)}\\\\
(a^3+b^3)=(a+b)(a^2-ab+b^2+3ab-3ab)\\\\
(a^3+b^3)=(a+b)(a^2+2ab+b^2-3ab)\\\\
(a^3+b^3)=(a+b)((a+b)^2-3ab)\\\\\\
But \;b=\frac{1}{a} \qquad and \;\;a+b=6\\\\
(a^3+\frac{1}{a}^3)=(6)((6)^2-3a*\frac{1}{a})\\\\
(a^3+\frac{1}{a}^3)=(6)(36-3)\\\\
(a^3+\frac{1}{a}^3)=(6)(33)\\\\
(a^3+\frac{1}{a}^3)=198\\\\$$

Melody  Feb 1, 2015
 #4
avatar+85644 
0

Compared to Melody and heureka.....I made that WAY too complicated...!!!

But, as Melody has said, "All roads lead to Rome!!!

 

CPhill  Feb 1, 2015
 #5
avatar+92196 
0

That is true Chris, that is soooo true   LOL

Melody  Feb 1, 2015

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