Question

Find the value of

\sqrt[3]{50+19\sqrt{7}}+\sqrt[3]{50-19\sqrt{7}}


Collected in the board: Cube root

Steven Zheng posted 1 week ago

Answer 1

Let

a = \sqrt[3]{50+19\sqrt{7} }
(1)
b = \sqrt[3]{50-10\sqrt{7} }
(2)

Addition of the equations after raising a and b to the 3rd power gives

a^3+b^3 = 100
(3)

multiplying (1) with (2) gives

ab = \sqrt[3]{50^2-(19\sqrt{7} )^2} =\sqrt[3]{-27} = -3
(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+9x-100 = 0
(7)

By observation, x = 4 is one of the solution for equation (7)

Then, divide the polynomial x^3+9x-100 by x-4 to find another factor

(x-4)(x^2+4x+25) = 0
(8)

Since x^2+4x+25 = (x+2)^2 +21 >0 , there is no real solution for the equation x^2+4x+25 = 0

Therefore, there's only one solution x = 4. Now we conclude

\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}} = 4

\sqrt[3]{50+19\sqrt{7}}+\sqrt[3]{50-19\sqrt{7}}=4

Steven Zheng posted 1 week ago

Answer 2

Transform the binomial with square root to the form of cube of binomial

50+19\sqrt{7}=(2+\sqrt{7})^3

50-19\sqrt{7}=(2-\sqrt{7})^3

Therefore

\sqrt[3]{50+19\sqrt{7}}+\sqrt[3]{50-19\sqrt{7}}=4

Steven Zheng posted 1 week ago

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