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What is x^3 + 1/8 when completely factored.

In (a+b)(a^2+2ab+b^2) form

Guest Nov 21, 2014

Best Answer 

 #1
avatar+20025 
+5

What is x^3 + 1/8 when completely factored

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

heureka  Nov 21, 2014
 #1
avatar+20025 
+5
Best Answer

What is x^3 + 1/8 when completely factored

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

heureka  Nov 21, 2014

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