Question

Factorize $27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz$ using identity

Answer 1

Hint: To solve this question we need the following identity:
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$

Complete step by step answer:
We have the given expression $27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz$. Now we need to make it similar to the expression ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ …(i) for this we have-
${{\left( 3x \right)}^{3}}+{{y}^{3}}+{{z}^{3}}-3.(3x).y.z$

Comparing this equation with equation (i) we have, $a=3x\ ,\ b=y\ ,\ c=z$

Hence, we can write the given expression as-

$\begin{align}

  & 27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz=(3x+y+z)\{{{(3x)}^{2}}+{{y}^{2}}+{{z}^{2}}-3xy-yz-3zx\} \\

 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

=(3x+y+z)(9{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3xy-yz-3zx) \\

\end{align}$

Note: If we don’t remember the identity ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$ it will be very difficult to solve this question. Someone could have tried to expand it using the formula of

${{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}+{{b}^{2}}-ab)$ but it would have taken us nowhere. All of these are standard formulas and must be remembered thoroughly.

 One must remember formulas regarding two to three variables and their squares and cubes. All of these formulas make solving equations quite easy and convenient otherwise it becomes very difficult to solve. We can try to solve this any other way and see that it has become very cumbersome.