+0

# Evaluate if .

0
266
5

Evaluate  if .

Guest Jan 31, 2015

#3
+92215
+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
Sort:

#1
+85726
+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
+19207
+10

Evaluate

if

.

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

heureka  Feb 1, 2015
#3
+92215
+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
#4
+85726
0

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

CPhill  Feb 1, 2015
#5
+92215
0

That is true Chris, that is soooo true   LOL

Melody  Feb 1, 2015

### 8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details