How to prove the equation is correct

I now the Lagrange identity: √a+√b=√a+√a2−b2+√a−√a2−b2\sqrt{a+\sqrt b}=\sqrt\frac{{a+\sqrt{a^2-b}}}{2}+\sqrt\frac{{a-\sqrt{a^2-b}}}{2}
but i didnt know how to prove that the equation
3√2+√5+3√2−√5=1\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=1
I do not think that should be used above identity. I tried to resolve the draw as irrational but appears more complicated, so please help me to solve.
Previously, thank you for the helping.




But 2−√5<02-\sqrt5<0 – lab bhattacharjee 2 days ago      The langrange identity only works when n=2n=2 in n√a+√b\sqrt[n]{a+\sqrt b}. In your problem, n=3n=3 so the langrange identity does not apply. In my post, I describe an alternative way to denest m√A+Bn√C\sqrt[m]{A+B\sqrt[n]{C}} without setting the expression equal to yy. – Frank 2 days ago ================= 5 Answers 5 ================= This screams Cardan's formula −q2=2,q24+p327=5 -\frac{q}{2}=2,\qquad \frac{q^2}{4}+\frac{p^3}{27}=5 Thus q=−4q=-4 and p=3p=3. What's the real root of the following equation? x3+3x−4=0 x^3+3x-4=0 I think you have meant 3√2+√5+3√2−√5\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5} Let y=3√2+√5+3√2−√5y=\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5} y3=2+√5+2−√5+3(3√2+√5)(3√2−√5)(3√2+√5+3√2−√5)=4+33√−1yy^3=2+\sqrt5+2-\sqrt5+3(\sqrt[3]{2+\sqrt5})(\sqrt[3]{2-\sqrt5})(\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5})=4+3\sqrt[3]{-1}y ⟺y3+3y−4=0\iff y^3+3y-4=0 whose only real root is 11      @dxiv, thanks for your observation. – lab bhattacharjee 2 days ago Proof: Let √a+√b=√x+√y\sqrt{a+\sqrt b}=\sqrt x+\sqrt y\tag1 Squaring both sides, we get a+√b=x+y+2√xya+\sqrt b=x+y+2\sqrt{xy}\tag2 And from (2)(2), we see that x+yx+y must equal aa, and 2√xy2\sqrt{xy} must equal √b\sqrt{b}. Thus, we have {x+y=axy=b4\begin{cases}x+y=a\\xy=\frac b4\end{cases} And using substitution x=a−yx=a-y, we get y(a−y)−b4=0⟹y2−ay+b4=0y(a-y)-\frac b4=0\implies y^2-ay+\frac b4=0 Solving should give you yy, and since x+y=ax+y=a, xx is the conjugate of yy. To prove 3√2+√5+3√2−√5\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}, another way is to set each nested radical equal to a+b√5a+b\sqrt5 and solve. Or more specifically, we have m√A+Bn√C=a+bn√C\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}\tag3and m√A−Bn√C=a−bn√C\sqrt[m]{A-B\sqrt[n]C}=a-b\sqrt[n]{C}\tag4 Observe that (1+√52)3=1+3√5+3⋅5+5√58=16+8√58=2+√5\left(\frac{1+\sqrt 5}{2}\right)^3=\frac{1+3\sqrt 5+3\cdot5+5\sqrt 5}{8} = \frac{16+8\sqrt 5}{8}=2+\sqrt 5 Similarly, (1−√52)3=1−3√5+3⋅5−5√58=16−8√58=2−√5\left(\frac{1-\sqrt 5}{2}\right)^3=\frac{1-3\sqrt 5+3\cdot5-5\sqrt 5}{8} = \frac{16-8\sqrt 5}{8}=2-\sqrt 5 Your equation transforms into \sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\sqrt[3]{\left(\frac{1+\sqrt 5}{2}\right)^3}+\sqrt[3]{\left(\frac{1-\sqrt 5}{2}\right)^3} = \frac{1+\sqrt 5}{2}+\frac{1-\sqrt 5}{2}=1\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\sqrt[3]{\left(\frac{1+\sqrt 5}{2}\right)^3}+\sqrt[3]{\left(\frac{1-\sqrt 5}{2}\right)^3} = \frac{1+\sqrt 5}{2}+\frac{1-\sqrt 5}{2}=1 Let a=\sqrt[3]{2+\sqrt 5}a=\sqrt[3]{2+\sqrt 5}, b=\sqrt[3]{2-\sqrt 5}b=\sqrt[3]{2-\sqrt 5}. By inspection a^3+b^3=4a^3+b^3=4 and a \cdot b=-1a \cdot b=-1. Let c=a+bc=a+b. Then: 4 = a^3+b^3=(a+b)(a^2 - a \cdot b + b^2) = (a+b)\left((a+b)^2 - 3 a \cdot b\right) = c^3 + 3 c 4 = a^3+b^3=(a+b)(a^2 - a \cdot b + b^2) = (a+b)\left((a+b)^2 - 3 a \cdot b\right) = c^3 + 3 c But the equation c^3 + 3 c - 4 = 0c^3 + 3 c - 4 = 0 has 11 as the unique real root, so c=1c=1.