What is the value of (2+root5)^1/3 + (2-root5)^1/3 ?

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Answer 1 · 2018-02-03T11:25:45

The value is $-2$

Let $x= (2+sqrt5)^(1/3)+(2-sqrt5)^(1/3)$ then

$x^3= {(2+sqrt5)^(1/3)+(2-sqrt5)^(1/3)}^3$

Reminder:

[ $(a+b)^3=a^3+b^3+3ab(a+b) , a^2-b^2=(a+b)(a-b)$ ]

and let $a=(2+sqrt5), b=(2-sqrt5):. ab=4-5=-1$

$:.x^3=(2+sqrt5)^(3*1/3)+(2-sqrt5)^(3*1/3)+3(2+sqrt5)(2-sqrt5)(2+sqrt5+2-sqrt5)$

or $x^3=2+cancel(sqrt5)+2-cancel(sqrt5)+3(4-5)(2+cancel(sqrt5)+2-cancel(sqrt5))$

or $x^3=4+3(-1)(4) or x^3 =4 -12 or x^3= -8$ or

$x = (-8)^(1/3) =-2$ . The value is $-2$ [Ans]