Chapter 2.5, Problem 65E

Solution

To find: The three cube roots of $8$ , $x^3=8$ .

The roots of the polynomial are $2,-1+\sqrt{3}i,-1-\sqrt{3}i$ .

Given information:

The given polynomial is $x^3=8$ .

Formula used:

  ${\left(a-b\right)}^3=\left(a-b\right)\left(a^2+ab+b^2\right)$.

The quadratic formula : $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculations:

The polynomial is $x^3=8.$

Rewrite the polynomial as $x^3-2^3=0\sin ce8=2^3$

Since which is difference of the cube so this can factorize as $\left(x-2\right)\left(x^2+2x+4\right)=0$ .

Now one linear factor and one quadratic factor are there.

  $\left(x-2\right)\left(x^2+2x+4\right)=0$

So each factor equate to $0$ to get value for $x.$

Thus ;

  $\begin{array}{c}x-2=0\\ x=2\end{array}$ so $x=2$ is the one real root for the polynomial $x^3=8$ .

Now find out the roots for the quadratic equation factor.

By using the quadratic formula, solve value for $x$ .

  $\begin{array}{c}x=\frac{-2\pm \sqrt{2^2-4\times 4}}{2}\\ =\frac{-2\pm \sqrt{4-4\times 4}}{2}\\ =\frac{-2\pm 2\sqrt{1-4}}{2}\\ =-1\pm \sqrt{-3}\\ =-1\pm \sqrt{3}i\end{array}$

So the no real roots of $x$ are $-1+\sqrt{3}i,-1-\sqrt{3}i.$

Hence the roots of the given polynomial are $2,-1+\sqrt{3}i,-1-\sqrt{3}i$ .