Question 1209219
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Answer: <font color=red>-5/27</font>


Explanation


{{{1/(u^3) + 1/(v^3)}}} turns into {{{(u^3+v^3)/( (uv)^3 )}}}


u^3 + v^3 = (u + v)(u^2 - uv + v^2) is the sum of cubes factoring rule


Use the identity u^2+v^2 = (u+v)^2-2uv to rewrite that previous equation like so
u^3 + v^3 = (u + v)(u^2 - uv + v^2)
u^3 + v^3 = (u + v)(u^2 + v^2 - uv)
u^3 + v^3 = (u + v)((u+v)^2-2uv - uv)
u^3 + v^3 = (u + v)((u+v)^2 - 3uv)


At this point we have terms involving u+v and uv


What can we do with this? We can use Vieta's Formulas.
But first 3x^2 + 5x + 7 = x^2 + 8x - 2 must be rearranged into 2x^2-3x+9 = 0
Then divide everything by the leading coefficient to get x^2-(3/2)x+9/2 = 0


Due to Vieta's formulas, the roots add to the negative of the x coefficient and multiply to the constant term. This applies only when the leading coefficient is 1.
We have these equations
u+v = 3/2
u*v = 9/2


So,
u^3 + v^3 = (u + v)((u+v)^2 - 3uv)
u^3 + v^3 = (3/2)*( (3/2)^2 - 3*(9/2) )
u^3 + v^3 = -135/8


And,
{{{1/(u^3) + 1/(v^3)}}} 


= {{{(u^3+v^3)/( (uv)^3 )}}}


= {{{(-135/8)/( (9/2)^3 )}}}


= {{{-5/27}}}


Therefore,
{{{1/(u^3) + 1/(v^3)=-5/27}}}
where u,v are the roots of 3x^2 + 5x + 7 = x^2 + 8x - 2


I used GeoGebra to verify the answer is correct.


-5/27 = -0.185185 approximately where the "185" repeats forever.


More practice is found <a href="https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1209218.html">here</a>
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