Question

Find the value of

\sqrt[3]{26+15\sqrt{3} } +\sqrt[3]{26-15\sqrt{3} }

Collected in the board: Cube root

Steven Zheng posted 1 week ago

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

Steven Zheng posted 1 week ago

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

Steven Zheng posted 1 week ago

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