Question 1098844
If we are not given the solution, we just need to solve
{{{(X+Y*2^(1/2))^3=7+5*2^(1/2)}}} .
If you like square roots better than fractional exponents,
feel free to write {{{2^"1 / 2"}}} as {{{sqrt(2)}}} .
I will have to write {{{1/2}}} as {{{"1/2"}}} in some places,
to make sure that you see the whole fractional exponent,
because I do not see the whole exponent in {{{2^(1/2)}}} .
{{{X^3+3*X^2*Y*2^(1/2)+3*X*(Y*2^(1/2))^2+(Y*2^(1/2))^3=7+5*2^(1/2)}}}
{{{X^3+3*X^2*Y*2^"1 / 2"+3*X*Y^2*2+Y*2^"3/2"=7+5*2^"1/2"}}}
{{{X^3+3*X^2*Y*2^"1/2"+6*X*Y^2+Y*2*2^"1/2"=7+5*2^"1/2"}}}
{{{X^3+6XY^2-7+(3X^2Y+2Y^3-5)*2^"1/2"=0}}}
Now we only need the one and only rational pair solution to
{{{system(X^3+6XY^2-7=0,3X^2+2Y^3-5=0)}}} .
When looking for rational solutions for polynomial equations,
we just try numbers that look promising.
Substituting {{{system(X=1,Y=1)}}} in those two equations,
we find that they are a solution.
Knowing that there was only one real solution to be found,
we know that {{{system(X=1,Y=1)}}} gives us
THE one and only answer, which is
{{{highlight(1+2^"1/2")}}} , also known as {{{1+sqrt(2)}}} .
 
NOTE: If the question was stated as "prove that {{{(7+5*2^(1/2))^(1/3) =1+2^(1/2)}}} ," all we would have needed to do is calculate
{{{(1+2^"1/2")^3}}} , by expanding that cube of a binomial,
and simplifying as needed.
As asked, the problem was not as difficult as it was intimidating,
encouraging you to give up.