Solving <mroot> x &#x2212;<!-- − --> 3 </mrow> 3 </mroot> +

Jordyn Calhoun

Jordyn Calhoun

Answered question

2022-05-22

Solving x 3 3 + 1 x 3 = 1

Answer & Explanation

vikafa4g

vikafa4g

Beginner2022-05-23Added 15 answers

Certainly for any real x
x 3 < x 1
so
x 3 3 < x 1 3
since cube root is a monotonically increasing function. So
x 3 3 + 1 x 3 = x 3 3 x 1 3 < 0 < 1.
Cooper Krause

Cooper Krause

Beginner2022-05-24Added 3 answers

We have that by A 3 B 3 = ( A B ) ( A 2 + A B + B 2 )
x 3 3 + 1 x 3 = x 3 3 x 1 3 =
= 2 ( x 3 ) 2 3 + ( x 3 ) ( x 1 ) 3 + ( x 1 ) 2 3 < 0
indeed the latter is true for x = 1 , 2 , 3 and for x 1 , 2 , 3 by AM-GM
( x 3 ) 2 3 + ( x 1 ) 2 3 > 2 | x 3 | | x 1 | 3 > 0
and therefore
( x 3 ) 2 3 + ( x 3 ) ( x 1 ) 3 + ( x 1 ) 2 3 > | x 3 | | x 1 | 3 > 0

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