How can I solve $1 + \frac{3}{2(x²-x+1)^{3/2}} = 5/2$

Question

I'm trying to solve this equation but I'm really stuck here, I know that I will have two solutions .

$$1 + \frac{3}{2(x²-x+1)^{3/2}} = \frac{5}{2}$$

I took $\frac{5}{2}$ to the other side and start solving it, I got to this point :

$$ 1 + \frac{3}{2(x²-x+1)^{3/2}} - \frac{5}{2} = 0$$

$$ \frac{2(x²-x+1)^{3/2} + 3 - 5 (x²-x+1)^{3/2}}{(x²-x+1)^{3/2}} = 0$$

$$ \frac{-3(x²-x+1)^{3/2} + 3 }{(x²-x+1)^{3/2}} = 0 $$

Which means:

$$ -3(x²-x+1)^{3/2} + 3 = 3 [-(x²-x+1)^{3/2} + 1] = 0 $$

I'm stuck here . I can't find the next steps to solve, I tried to use online tools to solve it, I got the results and the steps but it weren't so clear .

Please help me out on this equation and explain to me how can I solve it ? .

Answer 1

Hint:

Try moving the $1$ instead:

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

Then cross multiply to get:

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

Answer 2

$$1 + \frac{3}{2(x²-x+1)^{3/2}} = 5/2$$ $$ \frac{1}{(x²-x+1)^{3/2}} = 1$$ $$ (x²-x+1)^{3/2} = 1$$ $$ (x²-x+1) = 1$$

Answer 3

$$1+\frac{3}{\left(x^2-x+1\right)^{\frac{3}{2}}}=\frac{5}{2}\Longleftrightarrow x^2-x+1=\left(\frac{3}{\frac{5}{2}-1}\right)^{\frac{2}{3}}$$

Now, use the quadratic formula:

$$\text{a}x^2+\text{b}x+\text{c}=0\Longleftrightarrow x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}$$