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Let a and b be the solutions to 5x^2 - 11x + 4 = 4x^2 - 17x + 6. Find 1/a^3 + 1/b^3.

 Jul 2, 2024
 #1
avatar+1271 
+1

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! :)

 Jul 2, 2024
 #2
avatar+129739 
+1

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 

 

cool cool cool

 Jul 2, 2024
 #3
avatar+1271 
+1

Nice...Quite smart tactic here, CPhill. 

I loved the part where you did

\(1/a^3 + 1/b^3 = (a^3 + b^3) / (ab)^3 \\ \frac{(a + b) (a^2 + b^2 - ab)}{(ab)^3}\)

 

This is a very efficient way to complete the problem. 

Lol, I just brute forced it because I got bored...

 

At least we got that same answer!

 

~NTS

NotThatSmart  Jul 2, 2024

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