Question

1. ( x^3 + 6x^2 = 6x + 6 )

Explanation:

Step1: Rearrange the equation

Move all terms to one side to get a cubic equation: \(x^{3}+6x^{2}-6x - 6=0\)

Step2: Try rational roots

By Rational Root Theorem, possible rational roots are \(\pm1,\pm2,\pm3,\pm6\).
Test \(x = 1\): \(1 + 6-6 - 6=-5
eq0\)
Test \(x=-1\): \(-1 + 6 + 6-6 = 5
eq0\)
Test \(x = \sqrt{2}-2\) (or use numerical methods like Newton - Raphson). Let's use Newton - Raphson method. Let \(f(x)=x^{3}+6x^{2}-6x - 6\), \(f^{\prime}(x)=3x^{2}+12x - 6\)
Take initial guess \(x_0 = 1\), \(f(1)=-5\), \(f^{\prime}(1)=3 + 12-6 = 9\)
\(x_1=x_0-\frac{f(x_0)}{f^{\prime}(x_0)}=1-\frac{-5}{9}=\frac{14}{9}\approx1.5556\)
\(f(1.5556)\approx(1.5556)^{3}+6\times(1.5556)^{2}-6\times1.5556 - 6\approx3.75 + 14.58-9.33 - 6=2.99\)
\(f^{\prime}(1.5556)=3\times(1.5556)^{2}+12\times1.5556 - 6\approx7.31+18.67 - 6 = 19.98\)
\(x_2=x_1-\frac{f(x_1)}{f^{\prime}(x_1)}\approx1.5556-\frac{2.99}{19.98}\approx1.396\)
Continuing this process, we can find the real root approximately. Also, we can factor the cubic (though it may not factor nicely with rational factors). Alternatively, we can write the roots as:
The real root \(r\approx1.396\) and two complex roots from \(x^{2}+(6 + r)x+\frac{r^{3}+6r^{2}-6}{r}=0\) (using polynomial division or quadratic formula on the cubic after factoring out \((x - r)\))

Answer:

The roots of the equation \(x^{3}+6x^{2}-6x - 6 = 0\) are approximately \(x\approx1.396\) and two complex roots (found by quadratic formula on the depressed cubic after factoring out the real root). If we want an exact form, we can use Cardano's method, but the approximate real root is \(x\approx1.396\)