Question
Find the value of \sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}
Answer 1
Let
a =\sqrt[3]{7+5\sqrt{2} }
(1)
b = \sqrt[3]{7-5\sqrt{2} }
(2)
Addition of the equations after raising a and b to the 3rd power gives
a^3+b^3 = 14
(3)
multiplying (1) with (2) gives
ab = \sqrt[3]{7^2-(5\sqrt{2} )^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-14 = 0
(7)
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
(x-2)(x^2+2x+7) = 0
(8)
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
Answer 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