Chapter 2.5, Problem 66E

Solution

To find: The three cube roots of $-64$ , $x^3+64$ .

The roots of the polynomial are $-4,2+\sqrt{3}i,2-\sqrt{3}i$ .

Given information:

The given polynomial is $x^3=-64$ .

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 $x^3=-64.$

Rewrite the polynomial as $x^3+4^3=0\sin ce64=4^3$ .

Since which is difference of the cube so this can factorize as $\left(x+4\right)\left(x^2-4x+64\right)=0$ .

Now one linear factor and one quadratic factor are there.

  $\left(x+4\right)\left(x^2-4x+16\right)=0$

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

Thus ;

  $\begin{array}{c}x+4=0\\ x=-4\end{array}$ so $x=-4$ is the one real root for the polynomial $x^3=-64$ .

Now find out the roots for the quadratic equation factor.

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

  $\begin{array}{c}x=\frac{4\pm \sqrt{4^2-16\times 4}}{2}\\ =\frac{4\pm \sqrt{16-64}}{2}\\ =\frac{4\pm 4\sqrt{1-4}}{2}\\ =2\pm 2\sqrt{-3}\\ =2\pm 2\sqrt{3}i\end{array}$

So the no real roots of $x$ are $2+2\sqrt{3}i,2-2\sqrt{3}i$ .

Hence the roots of the given polynomial are $-4,2+\sqrt{3}i,2-\sqrt{3}i$ .