Another solution: first of all I have found that 3-2-sqrt 2=( sqrt2-1)^2.. therefore the denominator becomes sqrt2-1.. 1/(sqrt-1)= sqrt2 +1.. then we have the cubic root of 63(sqrt2+1)-8, or of 63sqrt2+55.. If we suppose that this cubic root is a+b sqrt2, we can cube this binomial obtaining the system a^3+6ab^2=55; 3a^2b+2b^3=63.. the first one can be written as a(a^2+6b^2)=55..,55 can be written as 5*11 or 55*1..the only chance is a^2+6b^2=55 and a=1.. if we substitute we obtain b=3 and this couple satisfies the second equation too.. so we have 1+3sqrt2
Another solution: first of all I have found that 3-2-sqrt 2=( sqrt2-1)^2.. therefore the denominator becomes sqrt2-1.. 1/(sqrt-1)= sqrt2 +1.. then we have the cubic root of 63(sqrt2+1)-8, or of 63sqrt2+55.. If we suppose that this cubic root is a+b sqrt2, we can cube this binomial obtaining the system a^3+6ab^2=55; 3a^2b+2b^3=63.. the first one can be written as a(a^2+6b^2)=55..,55 can be written as 5*11 or 55*1..the only chance is a^2+6b^2=55 and a=1.. if we substitute we obtain b=3 and this couple satisfies the second equation too.. so we have 1+3sqrt2
my way too
Do you mean simplify or solve. You can only splve an equation.
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