Algebra

First, let's cobine all like terms and solve for x. Combing all like terms, we have

\(x^{2}+6x-2=0\)

Using the quadratic equation, we find that

\(x=\sqrt{11}-3\\ x=-\sqrt{11}-3\)

Now, it's time for the more tedious part of this problem. This is not efficient but it should work. 

We have the equation

\(\frac{1}{\left(\sqrt{11}-3\right)^3}+\frac{1}{\left(-\sqrt{11}-3\right)^3}\)

Taking the LCM on both sides to get a common denominator, we get that

\(\frac{-126-38\sqrt{11}}{\left(\sqrt{11}-3\right)^3\left(-3-\sqrt{11}\right)^3}+\frac{38\sqrt{11}-126}{\left(\sqrt{11}-3\right)^3\left(-3-\sqrt{11}\right)^3}=\frac{-252}{\left(\sqrt{11}-3\right)^3\left(-3-\sqrt{11}\right)^3}\)

Now, we simply have to simplfy the denominator of the two equatins. We get that

\(\left(\left(\sqrt{11}-3\right)\left(-3-\sqrt{11}\right)\right)^3=(-2)^3 = -8\)

Thus, we have the answer as -252/-8 = 31.5

So 31.5 is our answer. 

Thanks! :)

Thx, NTS....here's another way  { no square roots involved !!! }

x^2 + 6x - 2  = 0

In the form mx^2 + nx + p  = 0

Product of the roots ab = -n/m =   -2  

So 2ab = -4

Sum of the roots a + b =  p/m =  -6     square both sides

a^2 + b^2 + 2ab = 36

a^2 + b^2 - 4 =36

a^2 + b^2  = 40

1/a^3  + 1/b^3  =

(a^3 + b^3) / (ab)^3           Note  :  a^3 + b^3 = (a + b) (a^2 + b^2 - ab)

So we have

(a + b) ( a^2 + b^2 -ab)  / (ab)^3  =

(-6) (40 - -2) / (-8)   = 

(6) (42 / 8)  =   31.5