Question
Find the value of
\sqrt[3]{26+15\sqrt{3} } +\sqrt[3]{26-15\sqrt{3} }
Answer 1
Let
a =\sqrt[3]{26+15\sqrt{3} }
(1)
b = \sqrt[3]{26-15\sqrt{3} }
(2)
Addition of the equations after raising a and b to the 3rd power gives
a^3+b^3 = 52
(3)
multiplying (1) with (2) gives
ab = \sqrt[3]{26^2-(15\sqrt{3} )^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-52 = 0
(7)
By observation, x = 4 is one of the solution for equation (7)
Then, divide the polynomial x^3-3x-52 by x-4 to find another factor
(x-4)(x^2+4x+13) = 0
(8)
Since x^2+4x+13 = (x+2)^2 + 9>0 , there is no real solution for the equation x^2+4x+13 = 0
Therefore, there's only one solution x = 4. Now we conclude
\sqrt[3]{26+15\sqrt{3} } +\sqrt[3]{26-15\sqrt{3} } =4
Answer 2
\sqrt[3]{26+15\sqrt{3} } +\sqrt[3]{26-15\sqrt{3} }
Transform the square root binomial to the form of cube of binomial
26+15\sqrt{3} = \big( 2+\sqrt{3} \big) ^3
26-15\sqrt{3} = \big( 2-\sqrt{3} \big) ^3
Therefore,
\sqrt[3]{26+15\sqrt{3} } +\sqrt[3]{26-15\sqrt{3} }=4