Question
Find the value of
\sqrt[3]{77+20\sqrt{13} } +\sqrt[3]{77-20\sqrt{13} }
Answer 1
Let
a = \sqrt[3]{77+20\sqrt{13} }
(1)
b = \sqrt[3]{77-20\sqrt{13} }
(2)
Addition of the equations after raising a and b to the 3rd power gives
a^3+b^3 = 154
(3)
multiplying (1) with (2) gives
ab = \sqrt[3]{77^2-(20\sqrt{13} )^2} =\sqrt[3]{729} = 9
(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^3-27x-154 = 0
(7)
By observation, x = 7 is one of the solution for equation (7)
Then, divide the polynomial x^3-27x-154 by x-7 to find another factor
(x-7)(x^2+7x+22) = 0
(8)
Since x^2+7x+22 = (x+\dfrac{7}{2})^2 + \dfrac{39}{4} >0 , there is no real solution for the equation x^2+7x+22 = 0
Therefore, there's only one solution x = 7. Now we conclude
\sqrt[3]{77+20\sqrt{13} } +\sqrt[3]{77-20\sqrt{13} } = 7
Answer 2
Transform the square root binomial to the form of cube of binomial
77+20\sqrt{13} = \Big( \dfrac{7+\sqrt{13} }{2} \Big) ^3
77-20\sqrt{13} = \Big( \dfrac{7-\sqrt{13} }{2} \Big) ^3
Therefore,
\sqrt[3]{77+20\sqrt{13} } +\sqrt[3]{77-20\sqrt{13} } = 7