Tricky Algebra Question III

Consider $x$ , $y$ as: $\begin{aligned} x &= \sqrt[3]{7+5\sqrt2}=\sqrt a+\sqrt b \\ y &= \sqrt[3]{7-5\sqrt2}=\sqrt a-\sqrt b \\ \end{aligned}$ And we will obtain: $\begin{aligned} x^3+y^3 &= 14 \\ x+y &= 2\sqrt a \\ xy &= -1 \\ \end{aligned}$ By using the formula of $(x+y)^{3}$ : $\begin{aligned} (x+y)^{3} &= x^3+y^3+3xy(x+y) \\ 8a\sqrt a &= 14-6\sqrt a \\ \sqrt a &= \frac{7}{\ 4a+3\ } \\ \end{aligned}$ Since $\sqrt a$ is a rational number and $a$ is an integer, so we can let $a=n^2$ which $n$ is a natural number. $(n)(4n^{2}+3)=(1)(7)=(7)(1)$ It is quite obvious that $n=1$ is the only solution.

Substitute $n=1$ back to the equation above: $\begin{aligned} xy &= -1=a-b \\ b &= a+1 \\ b &= 2 \\ \end{aligned}$

At last, we will found that: $\sqrt[3]{7+5\sqrt2}=\sqrt 1+\sqrt 2$

Thus, the value of $a+b$ is equal to $\boxed 3$ .