Answer:
x={0,1,6}
Given:
![x^3+6x=7x^2](https://img.qammunity.org/2023/formulas/mathematics/college/s20mpc1vo7mcd2vw3haatonyao2eixdjza.png)
Let us first arrange the equation and equate it to 0
![\begin{gathered} x^3+6x=7x^2 \\ x^3-7x^2+6x=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/be2x5z8wwpz93ueeks0mdjqtup1q3c15oz.png)
Factor out x:
![x(x^2-7x^{}+6)=0](https://img.qammunity.org/2023/formulas/mathematics/college/6ear0sesezazr7hmrbm2ooqxxkhvcu2ery.png)
Then, factor out the quadratic equation inside the parenthesis using factoring:
![\begin{gathered} x(x^2-7x^{}+6)=0 \\ x^2-7x^{}+6=(x-6)(x-1) \\ \rightarrow x(x-6)(x-1)=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wvv4ods79xndgmyofc3lrd4tqm4wzuur5y.png)
Now, we will equate each factor to 0 to find the solution to the equation
![\begin{gathered} x(x-6)(x-1)=0 \\ x=0 \\ ----------- \\ x-6=0 \\ x=6 \\ ----------- \\ x-1=0 \\ x=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/asvt94xl9uvpulj8vljxca9g2y9hwwtcug.png)
Therefore, the solutions to the equation is x={0,1,6}