First, we need to find the solutions a, b, and c of the equation:
[tex]x^3-20x=x^2[/tex]
We can rewrite it as:
[tex]\begin{gathered} x^3-x^{2}-20x=0 \\ \\ x(x^{2}-x-20)=0 \\ \\ x=0\text{ or }x^{2}-x-20=0 \end{gathered}[/tex]
Thus, one of the solutions is a = 0.
To find the other solutions, we can use the quadratic formula. We obtain:
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^{2}-4(1)(-20)}}{2(1)} \\ \\ x=\frac{1\pm\sqrt[]{1+80}}{2} \\ \\ x=\frac{1\pm\sqrt[]{81}}{2} \\ \\ x=\frac{1\pm9}{2} \\ \\ b=\frac{1-9}{2}=-4 \\ \\ c=\frac{1+9}{2}=5 \end{gathered}[/tex]
Now, we need to find the value of the expression:
[tex]\mleft(a+b+c\mright)+abc[/tex]
Using the previous solutions, we obtain:
[tex]\mleft(0-4+5\mright)+0(-4)(5)=1+0=1[/tex]
Therefore, the answer is 1.
