Question

Factorize the given expression, \[{{a}^{3}}{{b}^{3}}+{{c}^{3}}\].

Answer 1

Find the basic expansion of \[{{a}^{3}}+{{b}^{3}}\]. Similarly find the expansion of the given expression \[{{a}^{3}}{{b}^{3}}+{{c}^{3}}\] and try to factorize it.

Complete step-by-step answer:
A difference of two perfect cubes \[\left( {{a}^{3}}+{{b}^{3}} \right)\] can be factorized into,
\[{{a}^{3}}+{{b}^{3}}=(a+b).\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]
Now let us open the bracket and simplify the expression for \[{{a}^{3}}+{{b}^{3}}\].
\[\begin{align}
  & {{a}^{3}}+{{b}^{3}}=(a+b).\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
 & ={{a}^{3}}-{{a}^{2}}b+a{{b}^{2}}+b{{a}^{2}}-a{{b}^{2}}+{{b}^{3}} \\
 & ={{a}^{3}}+{{b}^{3}}+\left( {{a}^{2}}b-{{a}^{2}}b \right)+\left( a{{b}^{2}}-a{{b}^{2}} \right) \\
 & ={{a}^{3}}+{{b}^{3}}+0+0 \\
 & ={{a}^{3}}+{{b}^{3}}. \\
\end{align}\]
Thus similarly for \[{{a}^{3}}{{b}^{3}}-{{c}^{3}}\] it will become,
\[{{a}^{3}}{{b}^{3}}+{{c}^{3}}=(ab+c).\left( {{a}^{2}}{{b}^{2}}-abc+{{c}^{2}} \right)\]
If we try to factorize \[\left( {{a}^{2}}{{b}^{2}}-abc+{{c}^{2}} \right)\], this multi-variable trinomial using trial and error method then the factorization fails.
Thus the factorize result of \[{{a}^{3}}{{b}^{3}}-{{c}^{3}}=(ab-c).\left( {{a}^{2}}{{b}^{2}}-abc+{{c}^{2}} \right)\].
Thus, \[{{a}^{3}}{{b}^{3}}+{{c}^{3}}=(ab-c).\left( {{a}^{2}}{{b}^{2}}-abc+{{c}^{2}} \right).\]

Note: Here we found the expansion with the help of the formula \[{{a}^{3}}+{{b}^{3}}\]. It is important that you know the basic formula like these to work out the problems. This is an important concept so remember the formula.