Question

Resolve$a^3-b^3+c^3+3abc$ into factors

Answer 1

Finding the factors:

The given expression can be rewritten as

$a^3+(-b^3)+c^3-3a(-b)c$

$$\begin{gathered} \Rightarrow a^3+(-b^3)+c^3-3a(-b)c \\ \Rightarrow [a+(-)b+c][a^2+{(-b)}^2+c^2-a(-b)-(-)bc-ac] \\ \mathbf{\Rightarrow }\mathbf{[}\mathit{a}\mathbf{-}\mathit{b}\mathbf{+}\mathit{c}\mathbf{]}\mathbf{[}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}^{\mathbf{2}}\mathbf{+}\mathit{a}\mathit{b}\mathbf{+}\mathit{b}\mathit{c}\mathbf{-}\mathit{a}\mathit{c}\mathbf{]} \end{gathered}$$

Thus,$a^3+(-b^3)+c^3-3a(-b)c$ $=\mathbf{[}\mathit{a}\mathbf{-}\mathit{b}\mathbf{+}\mathit{c}\mathbf{]}\mathbf{[}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}^{\mathbf{2}}\mathbf{+}\mathit{a}\mathit{b}\mathbf{+}\mathit{b}\mathit{c}\mathbf{-}\mathit{a}\mathit{c}\mathbf{]}$