Question

Factorize the following expression ${{(x-y-z)}^{3}}-{{x}^{3}}+{{y}^{3}}+{{z}^{3}}$.

Answer 1

Hint:We can write the term ${{\left( x-y-z \right)}^{3}}$ as ${{\left( x-\left( y+z \right) \right)}^{3}}$. Then, we can use the formula ${{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{b}^{3}} \right)-3ab\left( a-b \right)$ to convert ${{\left( x-\left( y+z \right) \right)}^{3}}$into ${{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( y+z \right)\left( x-y \right)\left( x-z \right)$.

Complete step by step answer:
Before proceeding with the question, we must know what is meant by factorization. Factorization is defined as the breaking of an entity like a number, a polynomial, etc, into another product, entity, or factors, which when multiplied together give the original number. We must also know the formula for ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$, we can also write the formula for ${{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{b}^{3}} \right)-3ab\left( a-b \right)$.
Here, we have been given to factorize ${{(x-y-z)}^{3}}-{{x}^{3}}+{{y}^{3}}+{{z}^{3}}$.
First of all, let us assume ${{(x-y-z)}^{3}}-{{x}^{3}}+{{y}^{3}}+{{z}^{3}}.....(i)$.
Now, we can write ${{\left( x-y-z \right)}^{3}}$ as ${{\left( x-\left( y+z \right) \right)}^{3}}$. Now, using the formula ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$, we can write ${{\left( x-\left( y+z \right) \right)}^{3}}$as,
$\Rightarrow {{x}^{3}}-{{\left( y+z \right)}^{3}}-3{{x}^{2}}\left( y+z \right)+3x{{\left( y+z \right)}^{2}}$
Splitting ${{\left( y+z \right)}^{3}}$using the formula ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$ we get:
$\Rightarrow {{x}^{3}}-\left( {{y}^{3}}+{{z}^{3}}+3{{y}^{2}}z+3y{{z}^{2}} \right)-3{{x}^{2}}\left( y+z \right)+3x{{\left( y+z \right)}^{2}}$
$\Rightarrow {{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( {{y}^{2}}z+y{{z}^{2}} \right)-3{{x}^{2}}\left( y+z \right)+3x{{\left( y+z \right)}^{2}}$
Taking out $3$ as common we get:
$\Rightarrow {{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( yz\left( y+z \right)+{{x}^{2}}\left( y+z \right)-x{{\left( y+z \right)}^{2}} \right)$
Now, taking out $\left( y+z \right)$ as common we get:
$\Rightarrow {{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( y+z \right)\left( yz+{{x}^{2}}-x\left( y+z \right) \right)$
$\Rightarrow {{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( y+z \right)\left( yz+{{x}^{2}}-xy-xz \right)$
$\Rightarrow {{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( y+z \right)\left( x\left( x-y \right)-z\left( x-y \right) \right)$
Taking out $\left( x-y \right)$ common we get:
$\Rightarrow {{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( y+z \right)\left( x-y \right)\left( x-z \right)$
Therefore, we can write ${{\left( x-y-z \right)}^{3}}={{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( y+z \right)\left( x-y \right)\left( x-z \right)$.
We have ${{(x-y-z)}^{3}}-{{x}^{3}}+{{y}^{3}}+{{z}^{3}}$ as equation (i). Now, replacing ${{\left( x-y-z \right)}^{3}}$ within equation (i) we get:
$\Rightarrow {{x}^{3}}-{{y}^{3}}-{{z}^{3}}-3\left( y+z \right)\left( x-y \right)\left( x-z \right)-{{x}^{3}}+{{y}^{3}}+{{z}^{3}}$
After a cancellation we get:
$\Rightarrow -3\left( y+z \right)\left( x-y \right)\left( x-z \right)$
Therefore, after factorization of ${{(x-y-z)}^{3}}-{{x}^{3}}+{{y}^{3}}+{{z}^{3}}$, the answer is $-3\left( y+z \right)\left( x-y \right)\left( x-z \right)$.
Hence, ${{(x-y-z)}^{3}}-{{x}^{3}}+{{y}^{3}}+{{z}^{3}}$ factorise into $-3\left( y+z \right)\left( x-y \right)\left( x-z \right)$.

Note: Writing ${{\left( x-y-z \right)}^{3}}$ as ${{\left( x-\left( y+z \right) \right)}^{3}}$ will make it easy to solve the question. If one does not recollect the formula for ${{\left( a+b+c \right)}^{3}}$, which is a bit lengthy, then this approach can be used. One must be very careful with the calculations, especially with the signs while opening the bracket and taking out common terms. After seeing the problems of factorization, always try to take the common terms wherever possible and try to simplify the problem before applying all the formulas related to the problems. Remember that this formula ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$ can also be written as ${{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{b}^{3}} \right)-3ab\left( a-b \right)$. So, one must remember both forms of this formula.