Question 76096
<pre><font size = 4><b>
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}}} 

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It is well known that a solution to the cubic equation

{{{x^3+px+q=0}}} is given by the formula

{{{x=(-q/2 + sqrt(q^2/4 + p^3/27))^(1/3)+(-q/2-sqrt(q^2/4 + p^3/27))^(1/3))}}}

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/2 = 20}}} 

{{{sqrt(q^2/4 + p^3/27) = 14sqrt(2)}}}

The first gives q = -40

Substituting that in the second equation:

{{{sqrt((-40)^2/4 + p^3/27) = 14sqrt(2)}}}

{{{sqrt(400 + p^3/27) = 14sqrt(2)}}}

Squaring both sides:

{{{400 + p^3/27 = 196(2)}}}

{{{400 + p^3/27 = 392}}}

{{{p^3/27 = -8}}}

Multiply both sides by 27

{{{p^3 = -216}}}

Taking cube roots of both sides:

{{{p = -6}}}

So your expression is a solution to the cubic
equation  {{{x^3+px+q=0}}} or

{{{x^3-6x-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^3-6x-40=0}}}
{{{4^3-6(4)-40=0}}}
   {{{64-24-40=0}}}
          {{{0=0}}}

Since the cubic equation {{{x^3-6x-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</pre>