SOLUTION: This problem is from my daughter's (Arundhati's) VIII standard text book on advanced mathematics. I checked it through a calculator, the expressions comes to 4, but how to solve it

Algebra ->  Exponents -> SOLUTION: This problem is from my daughter's (Arundhati's) VIII standard text book on advanced mathematics. I checked it through a calculator, the expressions comes to 4, but how to solve it      Log On


   



Question 76096: This problem is from my daughter's (Arundhati's) VIII standard text book on advanced mathematics. I checked it through a calculator, the expressions comes to 4, but how to solve it:
Prove that
(20+14*(sqrt(2)))^(1/3) + (20-14*(sqrt(2)))^(1/3) = 4
MK Yadava

Answer by Edwin McCravy(20077) About Me  (Show Source):
You can put this solution on YOUR website!

This problem is from my daughter's (Arundhati's) VIII 
standard text book on advanced mathematics. I checked 
it through a calculator, the expressions comes to 4, 
but how to solve it: 

Prove that 
 

-----------------------------

It is well known that a solution to the cubic equation

x%5E3%2Bpx%2Bq=0 is given by the formula



Since your expression is of this form, let's find the 
values of p and q that would make this equivalent to 
your expression. To do this we equate like parts

-q%2F2+=+20 

sqrt%28q%5E2%2F4+%2B+p%5E3%2F27%29+=+14sqrt%282%29

The first gives q = -40

Substituting that in the second equation:

sqrt%28%28-40%29%5E2%2F4+%2B+p%5E3%2F27%29+=+14sqrt%282%29

sqrt%28400+%2B+p%5E3%2F27%29+=+14sqrt%282%29

Squaring both sides:

400+%2B+p%5E3%2F27+=+196%282%29

400+%2B+p%5E3%2F27+=+392

p%5E3%2F27+=+-8

Multiply both sides by 27

p%5E3+=+-216

Taking cube roots of both sides:

p+=+-6

So your expression is a solution to the cubic
equation  x%5E3%2Bpx%2Bq=0 or

x%5E3-6x-40=0

By DesCartes rule of signs, this cubic has
just one positive solution, so your expression
must be it.

Direct substitution of 4 into this equation shows
that 4 is its only positive solution:

  x%5E3-6x-40=0
4%5E3-6%284%29-40=0
   64-24-40=0
          0=0

Since the cubic equation x%5E3-6x-40=0 can have but 
one positive solution, and both your expression and 4 
must be this positive solution, your expression must
equal 4.

Edwin