Question
Find the value of
\sqrt[3]{26+15\sqrt{3} } +\sqrt[3]{26-15\sqrt{3} }
Find the value of
\sqrt[3]{26+15\sqrt{3} } +\sqrt[3]{26-15\sqrt{3} }
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^3-3x-52 by x-4 to find another factor
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
\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