What is x^3 + 1/8 when completely factored
$$x^3+\frac{1}{8}\\ \\
=x^3+\frac{1}{2^3}\\ \\
=x^3+ ( \frac{1}{2} ) ^3 \quad | \quad \textcolor[rgb]{1,0,0}{a^3+b^3 = (a+b)(a^2-ab+b^2) } \\\\
a = x \text{ and } b = \frac{1}{2} \\\\
x^3+ ( \frac{1}{2} ) ^3 = (x+\frac{1}{2})(x^2-x*\frac{1}{2}+(\frac{1}{2} )^2 )$$
What is x^3 + 1/8 when completely factored
$$x^3+\frac{1}{8}\\ \\
=x^3+\frac{1}{2^3}\\ \\
=x^3+ ( \frac{1}{2} ) ^3 \quad | \quad \textcolor[rgb]{1,0,0}{a^3+b^3 = (a+b)(a^2-ab+b^2) } \\\\
a = x \text{ and } b = \frac{1}{2} \\\\
x^3+ ( \frac{1}{2} ) ^3 = (x+\frac{1}{2})(x^2-x*\frac{1}{2}+(\frac{1}{2} )^2 )$$