Solve the equation without using square both sides Solve the equation without using square both sides:

$$(x+1)(2x^{3}-5)^{1/2}= 2x^{3}- 3x^{2}+x-2$$

I try to use Wolfram Alpha and here is the the answer:

https://www.wolframalpha.com/input/?i=(x%2B1)+(2x%5E3-+5)%5E(1%2F2)%3D+2x%5E3-3x%5E2%2Bx-2

I can' t continue. Help me!
The solution is $x=3$ and can be found as follows:

$2 x^3-5$ must be perfect square among cubic of few numbers we only have:

$2\times 3^3-5=54-5=49=7^2$

We check this:

$(3+1)\times 7= 28$

$2\times3^3-3\times 3^2+3-2=28$ 

Now suppose $(2x^3-5)^{1/2}=a$ then we have:

$a(x+1)=a^2-3x^2+x+3$.. or $a^2-(x+1)a-3x^2+x+3=0$

The condition for having solution in R is:

$\Delta=13x^2-2x-11=m^2; m ∈ R$

$2x^3-5=k^2; k∈ R$

The solution of question leads to the solution of this Diophantine system of equations.One solution is $x=3,..k=7,..m=10$
The solution is $x=3$ and can be found as follows:

$2 x^3-5$ must be perfect square among cubic of few numbers we only have:

$2\times 3^3-5=54-5=49=7^2$

We check this:

$(3+1)\times 7= 28$

$2\times3^3-3\times 3^2+3-2=28$ 

Now suppose $(2x^3-5)^{1/2}=a$ then we have:

$a(x+1)=a^2-3x^2+x+3$.. or $a^2-(x+1)a-3x^2+x+3=0$

The condition for having solution in R is:

$\Delta=13x^2-2x-11=m^2; m ∈ R$

$2x^3-5=k^2; k∈ R$

The solution of question leads to the solution of this Diophantine system of equations.One solution is $x=3,..k=7,..m=10$