Mathematico | Mathematico

The solutions of the equation

(x^{2} + 2 x + 1)^{2} - 3 (x + 1)^{2} - 4 = 0, x \in R

are $α$ and $β$ . Find the value of

α^{3} + β^{3}

Hint

Note that

x^{2} + 2 x + 1 = (x + 1)^{2}

Hint

So the equation becomes

{((x + 1)^{2})}^{2} - 3 (x + 1)^{2} - 4 = 0

Hint

Let

u = (x + 1)^{2}

and solve

Solution

Notice that $x^{2} + 2 x + 1 = (x + 1)^{2}$ , so

\begin{aligned} (x^{2} + 2 x + 1)^{2} - 3 (x + 1)^{2} - 4 & = 0 \\ (x + 1)^{4} - 3 (x + 1)^{2} - 4 & = 0 \end{aligned}

Now let $u = (x + 1)^{2}$

\begin{aligned} u^{2} - 3 u - 4 & = 0 \\ (x - 4) (u + 1) & = 0 \\ u & \in {- 1, 4} \end{aligned}

Notice that $(x + 1)^{2} = - 1$ is not possible (a real number squared cannot be negative), so we only need to consider

\begin{aligned} (x + 1)^{2} & = 4 \\ x + 1 & = \pm 2 \\ x & = 1 \pm 2 \\ x & \in {- 1, 3} \end{aligned}