SOLUTION: If 4b² + 1/b² = 2, find 8b³ + 1/b³

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Question 1142422: If 4b² + 1/b² = 2, find 8b³ + 1/b³
Found 3 solutions by greenestamps, ikleyn, MathTherapy:
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


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NOTE: IGNORE THIS SOLUTION!

The method was sound but the math was defective. I absent-mindedly added 2 to both sides of the original equation instead of the required 4.

See the response from tutor @ikleyn for the correct solution.

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This is a common type of question on competitive math tests at the high school level.

4b%5E2%2B1%2Fb%5E2+=+2

Note that both terms on the left are perfect squares:

%282b%29%5E2+%2B+%281%2Fb%29%5E2+=+2

Now remember that in squaring a binomial x+y, you get two perfect square terms plus a middle term: %28x%2By%29%5E2+=+x%5E2%2B2xy%2By%5E2.

Then note that if the binomial is of the form x+1/x, the middle term in the square is a constant: %28x%2B1%2Fx%29%5E2+=+x%5E2%2B2%2B1%2Fx%5E2

So, in this problem, add 2 to both sides; the left side becomes a perfect square trinomial:

%282b%29%5E2+%2B+4+%2B+%281%2Fb%29%5E2+=+4
%282b%2B1%2Fb%29%5E2+=+4
2b%2B1%2Fb+=+2

Now note that, in the expression you are to evaluate, both terms are perfect cubes: 8b^3 = (2b)^3; 1/b^3 = (1/b)^3.

You will get both of those terms, plus other terms, when you cube (2b+1/b):



This expression can be simplified using the value you now know of 2b+1/b:



So now you have

%282b%2B1%2Fb%29%5E3+=+2%5E3+=+8+=+%288b%5E3%2B1%2Fb%5E3%29%2B12

and so

8b%5E3%2B1%2Fb%5E3+=+8-12+=+-4

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
If 4b² + 1/b² = 2, find 8b³ + 1/b³
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            The solution by the other tutor has errors.

            So I came to provide a correct solution. 


You are given  


    4b%5E2 + 1%2Fb%5E2 = 2.      (1)


It is the same as


    %282b%29%5E2 + 1%2Fb%5E2 = 2.


Add 4 to both sides. You will get


    %282b%29%5E2 + 4 + b%5E2 = 6.      (2)


The left side is  nothing else as  %282b+%2B+1%2Fb%29%5E2.  Therefore, the equation (2) takes the form


    %282b+%2B+1%2Fb%29%5E2 = 6.


Take the square root from both sides of the last equation. You will get


    2b+%2B+1%2Fb = +/- sqrt%286%29.     (3)


Now, 


    %282b++%2B+1%2Fb%29%5E3 = %282b%29%5E3 + 3%2A%282b%29%5E2%2A%281%2Fb%29 + 3%2A%282b%29%2A%281%2Fb%5E2%29 + 1%2Fb%5E3 = 8b%5E3 + 3%2A4%2Ab%5E2%2A%281%2Fb%29 + 3%2A%282b%29%2A%281%2Fb%5E2%29 + 1%2Fb%5E3 = 8b%5E3 + 12b + 6%2Fb + 1%2Fb%5E3 = 8b%5E3 + 6%2A%282b+%2B+1%2Fb%29 + 1%2Fb%5E3.


Thus the very first part of this chain of equalities is equal to its very last part


    %282b++%2B+1%2Fb%29%5E3 = 8b%5E3 + 6%2A%282b+%2B+1%2Fb%29 + 1%2Fb%5E3.


Hence, 


    8b%5E3 + 1%2Fb%5E3 = %282b++%2B+1%2Fb%29%5E3 - 6%2A%282b+%2B+1%2Fb%29.


Now the final step is to substitute expression (3) into the last equality.


Since expression (3) has the sign +/-, I will make this substitution in two lines.



    Case 1.  If  2b+%2B+1%2Fb = + sqrt%286%29,  then  8b%5E3 + 1%2Fb%5E3 = %28sqrt%286%29%29%5E3 - 6%2Asqrt%286%29 = 6%2Asqrt%286%29 - 6%2Asqrt%286%29 = 0.



    Case 2.  If  2b+%2B+1%2Fb = - sqrt%286%29,  then  8b%5E3 + 1%2Fb%5E3 = %28-sqrt%286%29%29%5E3 - 6%2A%28-sqrt%286%29%29 = -6%2Asqrt%286%29 + 6%2Asqrt%286%29 = 0.



You see that for any of the two cases the answer is 0 (zero, ZERO).



ANSWER.  If  4b%5E2 + 1%2Fb%5E2 = 2,  then   8b%5E3 + 1%2Fb%5E3 = 0.

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This result seems to be  VERY  INEXPECTED,  and you might be overwhelmed / stunned by this answer,

but the deep reason why it is so is that the given equation  HAS  NO  real  roots  "b".

It has  ONLY  COMPLEX  NUMBER  solutions  (!)


See the attached plot of the function   y = 2x%5E2 + 1%2Fx%5E2. 


    graph%28+330%2C+330%2C+-5%2C+5%2C+-1%2C+10%2C%0D%0A++++++++++2x%5E2+%2B+1%2Fx%5E2%0D%0A%29


          Plot y = 2x%5E2 + 1%2Fx%5E2



The plot shows that the function  y = 2x%5E2 + 1%2Fx%5E2  is always greater than 2, so the given equation has no real roots.


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This problem is slightly  ABOVE  an averaged  High school Math competition level,  since it contains  TWO  UNDERWATER  STONES
instead of standard  ONE.

The second  underwater stone  is the complexity of the roots to the given equation,  which makes it difficult to a student
to believe that the obtained result / (answer)  is correct.


For the standard problems / (exercises) of the  High school competition level,  see the lessons
    - HOW TO evaluate expressions involving  %28x+%2B+1%2Fx%29,  %28x%5E2%2B1%2Fx%5E2%29  and  %28x%5E3%2B1%2Fx%5E3%29
    - Advanced lesson on evaluating expressions
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Evaluation, substitution".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
If 4b² + 1/b² = 2, find 8b³ + 1/b³
matrix%281%2C3%2C+4b%5E2+%2B+1%2Fb%5E2%2C+%22=%22%2C+2%29

matrix%281%2C3%2C+2b%284b%5E2+%2B+1%2Fb%5E2%29%2C+%22=%22%2C+2b%282%29%29 ------ Multiplying each side by 2b

  



matrix%281%2C3%2C+%281%2Fb%29%284b%5E2+%2B+1%2Fb%5E2%29%2C+%22=%22%2C+%281%2Fb%292%29 --- Multiplying each side by 1%2Fb