can someone help me on this?
solve the quadratic equations by factoring
(x + 2)^3 = x^3 + 8
If you know 8 = 2³, you see a sum of squares.
$\displaystyle
x^3 + 2^3 = \left( {x + 2} \right)\left( {x^2 - 2x + 4} \right)
$
Everything to the left, factor (x+2) and simplify:
$\displaystyle
\begin{array}{l}
\left( {x + 2} \right)^3 - \left( {x + 2} \right)\left( {x^2 - 2x + 4} \right) = 0 \\ \\
\left( {x + 2} \right)\left( {\left( {x + 2} \right)^2 - \left( {x^2 - 2x + 4} \right)} \right) = 0 \\ \\
\left( {x + 2} \right)\left( {6x} \right) = 0 \Leftrightarrow x = - 2 \vee x = 0 \\
\end{array}
$
Or, less elegantly:Originally Posted by danielwu
$\displaystyle (x+2)^3 = x^3 + 3 \cdot 2x^2 + 3 \cdot 2^2x + 2^3 = x^3 + 6x^2 + 12x + 8$
So
$\displaystyle x^3 + 6x^2 + 12x + 8 = x^3 + 8$
$\displaystyle 6x^2 + 12 x = 0$
$\displaystyle (6x)(x+2) = 0$
etc.
-Dan