Question
Find the value of
\sqrt[3]{50+19\sqrt{7}}+\sqrt[3]{50-19\sqrt{7}}
Find the value of
\sqrt[3]{50+19\sqrt{7}}+\sqrt[3]{50-19\sqrt{7}}
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 = 4 is one of the solution for equation (7)
Then, divide the polynomial x^3+9x-100 by x-4 to find another factor
Since x^2+4x+25 = (x+2)^2 +21 >0 , there is no real solution for the equation x^2+4x+25 = 0
Therefore, there's only one solution x = 4. Now we conclude
\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}} = 4
\sqrt[3]{50+19\sqrt{7}}+\sqrt[3]{50-19\sqrt{7}}=4
Transform the binomial with square root to the form of cube of binomial
50+19\sqrt{7}=(2+\sqrt{7})^3
50-19\sqrt{7}=(2-\sqrt{7})^3
Therefore
\sqrt[3]{50+19\sqrt{7}}+\sqrt[3]{50-19\sqrt{7}}=4