Question

solve for x.\\(x^3 = 8\\)\
write your answers in simplified, rationalized form.

Explanation:

Step1: Take cube root of both sides

To solve \( x^3 = 8 \), we take the cube root of both sides. The cube root of a number \( a \) is a number \( b \) such that \( b^3=a \). So we have \( \sqrt[3]{x^3}=\sqrt[3]{8} \).

Step2: Simplify both sides

Simplifying the left side, \( \sqrt[3]{x^3}=x \) (by the property of cube roots \( \sqrt[3]{y^3}=y \) for any real number \( y \)). For the right side, we know that \( 2^3 = 8 \), so \( \sqrt[3]{8}=2 \). Thus, \( x = 2 \). Also, for the complex roots, we can use the formula for the roots of a cubic equation \( x^3 - 8=0 \), which can be factored as \( (x - 2)(x^2+2x + 4)=0 \). Solving \( x^2+2x + 4 = 0 \) using the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 2 \), and \( c = 4 \). We get \( x=\frac{-2\pm\sqrt{4 - 16}}{2}=\frac{-2\pm\sqrt{- 12}}{2}=\frac{-2\pm2i\sqrt{3}}{2}=-1\pm i\sqrt{3} \). But since the problem might be expecting real and complex roots (as the interface has multiple boxes), we list all three roots.

Answer:

\( 2 \), \( -1 + i\sqrt{3} \), \( -1 - i\sqrt{3} \)