How can I solve the equation 8a³-1=4a+6a¹/³+1 in the real number system?

[anemone]

Solve the equation below in the real number system:

$8a^3-1=4a+\sqrt[3]{6a+1}$

 

[Wilmer]

Step1: 6a + 1 = (8a^3 - 4a - 1)^3 :)

 

[MarkFL]

Step1: 6a + 1 = (8a^3 - 4a - 1)^3 :)

https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcSG5Rdg-yNUGpPJVy3JqYQ5GBBaMKN0wJI57ajtnfRd6XsVXHz5

(Shake) Wilmer...http://mathhelpboards.com/challenge-questions-puzzles-28/guidelines-posting-answering-challenging-problem-puzzle-3875.html

 

[anemone]

(Shake) Wilmer...http://mathhelpboards.com/challenge-questions-puzzles-28/guidelines-posting-answering-challenging-problem-puzzle-3875.html

Aww MarkFL, don't be so harsh to your friend, hehehe...

 

[MarkFL]

Aww MarkFL, don't be so harsh to your friend, hehehe...

Okay, now we have 2 in the corner...anyone else? (Giggle)

 

[Wilmer]

1000 apologies of which you may have one!

Seriously, forgot the rules.

 

[anemone]

Solve the equation below in the real number system:

$8a^3-1=4a+\sqrt[3]{6a+1}$

Hint:

Spoiler

Injective Function

 

[anemone]

Solution of other:

Spoiler

$8a^3-1=4a+\sqrt[3]{6a+1}$ can be rewritten to become

$(2a)^3+2a=6a+1+\sqrt[3]{6a+1}$

but the function $f(x)=x^3+x$ is an injective function, therefore we deduce that

$(2a)^3=6a+1$

$8a^3-6a=1$

$4a^3-3a=\dfrac{1}{2}$

Note that

1. When $a<-1$, we have $4a^3-3a<-1$ and

2. When $a>1$, we have $4a^3-3a>1$.

So we can let $a=\cos x$ and this transform the equation $4a^3-3a=\dfrac{1}{2}$ into $\cos 3x=\dfrac{1}{2}$, solve this for $x$ we get:

$x=\dfrac{\pi}{9}+\dfrac{2k\pi}{3},\,k=0,\,1,\,2,\,\cdots$

Back substitute the values for $x$ into $a=\cos x$, the solutions of the given equation $8a^3-1=4a+\sqrt[3]{6a+1}$ are $a_1=\cos \dfrac{\pi}{9}$, $a_2=\cos \dfrac{7\pi}{9}$ and $a_3=\cos \dfrac{13\pi}{9}$.