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# Algebra

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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
+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

Thanks! :)

Jul 2, 2024
#2
+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

Jul 2, 2024
#3
+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