$$\boxed{(a^3+b^3)=(a+b)(a^2-ab+b^2)}\\\\
(a^3+b^3)=(a+b)(a^2-ab+b^2+3ab-3ab)\\\\
(a^3+b^3)=(a+b)(a^2+2ab+b^2-3ab)\\\\
(a^3+b^3)=(a+b)((a+b)^2-3ab)\\\\\\
But \;b=\frac{1}{a} \qquad and \;\;a+b=6\\\\
(a^3+\frac{1}{a}^3)=(6)((6)^2-3a*\frac{1}{a})\\\\
(a^3+\frac{1}{a}^3)=(6)(36-3)\\\\
(a^3+\frac{1}{a}^3)=(6)(33)\\\\
(a^3+\frac{1}{a}^3)=198\\\\$$
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......
Evaluate
if
$$\left(a+\frac{1}{a}\right)^3=6^3\\\\
a^3+3a^2*\frac{1}{a}+3a*\frac{1}{a^2}+\frac{1}{a^3}=6^3\\\\
a^3+3a+3\frac{1}{a}+\frac{1}{a^3}=6^3\\\\
a^3+\frac{1}{a^3}+3a+3\frac{1}{a}=6^3\\\\
a^3+\frac{1}{a^3}+3\underbrace{\left(a+\frac{1}{a}\right)}_{=6}=6^3\\\\
a^3+\frac{1}{a^3}+3*6=6^3\\\\
a^3+\frac{1}{a^3}=6^3-3*6\\\\
a^3+\frac{1}{a^3}=6(6^2-3)\\\\
a^3+\frac{1}{a^3}=6(33)\\\\
a^3+\frac{1}{a^3}=198$$
$$\boxed{(a^3+b^3)=(a+b)(a^2-ab+b^2)}\\\\
(a^3+b^3)=(a+b)(a^2-ab+b^2+3ab-3ab)\\\\
(a^3+b^3)=(a+b)(a^2+2ab+b^2-3ab)\\\\
(a^3+b^3)=(a+b)((a+b)^2-3ab)\\\\\\
But \;b=\frac{1}{a} \qquad and \;\;a+b=6\\\\
(a^3+\frac{1}{a}^3)=(6)((6)^2-3a*\frac{1}{a})\\\\
(a^3+\frac{1}{a}^3)=(6)(36-3)\\\\
(a^3+\frac{1}{a}^3)=(6)(33)\\\\
(a^3+\frac{1}{a}^3)=198\\\\$$