Question
Find the value of
\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}}
Answer 1
Let
a = \sqrt[3]{38+17\sqrt{5} }
(1)
b = \sqrt[3]{38-17\sqrt{5} }
(2)
Addition of the equations after raising a and b to the 3rd power gives
a^3+b^3 = 76
(3)
multiplying (1) with (2) gives
ab = \sqrt[3]{38^2-(17\sqrt{5} )^2} =\sqrt[3]{-1} = -1
(4)
Transform the identity of perfect square of binomial to
(a+b)^3 = a^3+b^3+3ab(a+b)
(5)
Let
x = a+b
(6)
Substituting (3), (4) and (6) into (5) results in a depressed cubic equation
x^3+3x-76 = 0
(7)
By observation, x = 4 is one of the solution for equation (7)
Then, divide the polynomial x^3+3x-76 by x-4 to find another factor
(x-4)(x^2+4x+19) = 0
(8)
Since x^2+4x+19 = (x+2)^2 +15 >0 , there is no real solution for the equation x^2+4x+19 = 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
Answer 2
Transform binomial with square root to the form of cube of binomial
38+17\sqrt{5}=(2+\sqrt{5})^3
38-17\sqrt{5}=(2-\sqrt{5})^3
Therefore
\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}}=4