Question
Find the value of \sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}
Find the value of \sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}
Let
Addition of the equations after raising a and b to the 3rd power gives
multiplying (1) with (2) gives
Transform the identity of perfect square of binomial to
Let
Substituting (3), (4) and (6) into (5) results in a depressed cubic equation
By observation, x = 2 is one of the solution for equation (7)
Then, divide the polynomial x^3+3x-14 by x-2 to find another factor
Since x^2+2x+7= (x+1)^2 + 6>0 , there is no real solution for the equation x^2+2x+7 = 0
Therefore, there's only one solution x = 2. Now we conclude
\sqrt[3]{7+5\sqrt{2} } +\sqrt[3]{7-5\sqrt{2} } =2
Transform the binomial with square root to the form of cube of binomial
7+5\sqrt{2}=(1+\sqrt{2})^3
7-5\sqrt{2}=(1-\sqrt{2})^3
Find cube root: \sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}
Find cube root: \sqrt[3]{7-5\sqrt{2}}=1-\sqrt{2}
Therefore
\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}=2