Question

Other Equations Find all real solutions of the equation.

$x^6-2x^3-3=0$

Short answer

The real solutions of the given equation are$x=-1$ and $x=\sqrt[3]{3}$.

Step-by-step solution

Step 1. Definition of equation.

Equation is a mathematical statement in which two expressions are equal to each other.

Step 2. Given Information.

The given equation is:

$x^6-2x^3-3=0$

Step 3. Simplify the given equation.

Rewrite the given equation.

${\left(x^3\right)}^2-2x^3-3=0$

Substitute$x^3=y$ in the above equation.

localid="1643899830200" $\begin{array}{c}{\left(y\right)}^2-2\left(y\right)-3=0\\ y^2-2y-3=0\end{array}$

Step 4. Solve the above equation.

In the above equation$2$ can be written as$3-1$ because the sum of$3$ and$-1$ is$2$ and their product is $-3$.

$\begin{array}{c}{y}^2-\left(3-1\right)y-3=0\\ y^2-3y+y-3=0\\ y(y-3)+(y-3)=0\\ (y+1)(y-3)=0\end{array}$

Use the zero product property.

role="math" localid="1643900134618" $$\begin{gathered} y+1=0 \\ y=-1 \\ {x}^3=-1\left[\because y=x^3\right] \\ x=\sqrt[3]{-1} \\ x=-1 \end{gathered}$$

And,

role="math" localid="1643900258481" $$\begin{gathered} y-3=0 \\ y=-1 \\ {x}^3=-1\left[\because y=x^3\right] \\ x=\sqrt[3]{-1} \\ x=-1 \end{gathered}$$

Thus, the real solutions of the given equation are$x=-1$ and $x=\sqrt[3]{3}$.