An algebra problem by abhishek alva

Algebra Level 3

2 + 5 3 + 2 5 3 = ? \large \sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5} =\, ?


The answer is 1.

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3 solutions

Chew-Seong Cheong
Oct 10, 2016

x = 2 + 5 3 + 2 5 3 x 3 = ( 2 + 5 ) + 3 ( 2 + 5 ) 2 3 ( 2 5 ) 1 3 + 3 ( 2 + 5 ) 1 3 ( 2 5 ) 2 3 + ( 2 5 ) = 4 + 3 ( 2 + 5 ) 1 3 ( 2 5 ) 1 3 [ ( 2 + 5 ) 1 3 + ( 2 5 ) 1 3 ] = 4 + 3 ( 4 5 ) 1 3 x = 4 3 x \begin{aligned} x & = \sqrt[3]{2+\sqrt 5} + \sqrt[3]{2-\sqrt 5} \\ x^3 & = \color{#3D99F6}{(2+\sqrt 5)} + 3(2+\sqrt 5)^\frac 23 (2-\sqrt 5)^\frac 13 + 3(2+\sqrt 5)^\frac 13 (2-\sqrt 5)^\frac 23 + \color{#3D99F6}{(2-\sqrt 5)} \\ & = \color{#3D99F6}{4} + 3\color{#D61F06}{(2+\sqrt 5)^\frac 13 (2-\sqrt 5)^\frac 13}\color{#20A900}{[(2+\sqrt 5)^\frac 13 + (2-\sqrt 5)^\frac 13]} \\ & = 4 + 3\color{#D61F06}{(4-5)^\frac 13}\color{#20A900}{x} \\ & = 4 - 3x \end{aligned}

x 3 + 3 x 4 = 0 ( x 1 ) ( x 2 + x + 4 ) = 0 x = 1 \begin{aligned} \implies x^3 + 3x - 4 & = 0 \\ (x-1)(x^2+x+4) & = 0 \\ \implies x & = \boxed{1} \end{aligned}

x 2 + x + 4 x^2+x+4 has no real root.

Siddharth Yadav
Oct 10, 2016

The R.H.S.of the question can be easily manipulated to give 1 as the answer. First of all consider: [2+5^(1/2)]^1/3= [5^(1/2) +1]/2 and ; [2-5^(1/2)]^1/3 =[1-5^(1/2) ]/2
Adding the above two equations clearly gives R.H.S.of the question I.e. x=1

Ayush G Rai
Oct 10, 2016

Let x = 2 + 5 3 + 2 5 3 . x=\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}.
x 2 + 5 3 2 5 3 = 0. x-\sqrt[3]{2+\sqrt5}-\sqrt[3]{2-\sqrt5}=0. Now let a = x , b = 2 + 5 3 a=x,b=-\sqrt[3]{2+\sqrt5} and c = 2 5 3 . c=-\sqrt[3]{2-\sqrt5}.
Since a + b + c = 0 ; a 3 + b 3 + c 3 = 3 a b c . a+b+c=0;a^3+b^3+c^3=3abc.
x 3 2 5 2 + 5 = 3 × x ( 2 + 5 ) ( 2 5 ) 3 = 3 x . x^3-2-\sqrt5-2+\sqrt5=3\times x\sqrt[3]{(2+\sqrt5)(2-\sqrt5)}=-3x.
Therefore, x 3 + 3 x 4 = 0. x^3+3x-4=0. The rational root of this equation is 1 . \boxed{1}.



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