SOLUTION: Let u and v be the solutions to 3x^2 + 5x + 7 = x^2 + 8x - 2. Find 1/u^3 + 1/v^3

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Let u and v be the solutions to 3x^2 + 5x + 7 = x^2 + 8x - 2. Find 1/u^3 + 1/v^3      Log On


   

Question 1209219: Let u and v be the solutions to 3x^2 + 5x + 7 = x^2 + 8x - 2. Find
1/u^3 + 1/v^3
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: -5/27

Explanation

1%2F%28u%5E3%29+%2B+1%2F%28v%5E3%29 turns into %28u%5E3%2Bv%5E3%29%2F%28+%28uv%29%5E3+%29

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%2F%28u%5E3%29+%2B+1%2F%28v%5E3%29

= %28u%5E3%2Bv%5E3%29%2F%28+%28uv%29%5E3+%29

= %28-135%2F8%29%2F%28+%289%2F2%29%5E3+%29

= -5%2F27

Therefore,
1%2F%28u%5E3%29+%2B+1%2F%28v%5E3%29=-5%2F27
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 here