Question

The factors of given algebraic expression \[{{x}^{3}}-1+{{y}^{3}}+3xy\] are:-
A). \[\left( x-1+y \right)\left( {{x}^{2}}+1+{{y}^{2}}+x\ +y-xy \right)\]
B). \[\left( x+1+y \right)\left( {{x}^{2}}+{{y}^{2}}+1-xy-x-y \right)\]
C). \[\left( x-1+y \right)\left( {{x}^{2}}-1-{{y}^{2}}+x+y+xy \right)\]
D). \[3\left( x+y-1 \right)\left( {{x}^{2}}+{{y}^{2}}-1 \right)\]

Answer 1

Hint:- Before solving this question, we must know about the different identities that are used for factorization of numbers. These identities help in solving polynomials by breaking them down into simpler polynomials and which later help in calculating numeric values easily. These identities mentioned below are known as the identities for the factorization of cubic polynomial:-
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\].
\[{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}\] .
\[{{a}^{3}}+\ {{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\].
\[{{a}^{3}}-{{b}^{3}}\ =\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\].
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)\].

Complete step-by-step solution -
Factorization means breaking down any polynomial or a number into a product of its factor, which on dividing that polynomial or number gives a real number.
For example: Factorization of 12 gives : $2\times 2\times 3$ and factorization of ${{x}^{2}}+4x+4$ gives ( x + 2 )( x + 2 ).
Let us now solve this question.
In the hint provided above, we can see that the identities for the factorization of cubic polynomials are provided. We are going to use one of those identities for the solution of this question.
As mentioned in the hint provided above, \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] is equal to \[\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)\] . So, for the solution of this question, we shall use the same identity.
Therefore, \[{{x}^{3}}-1+{{y}^{3}}+3xy\] can also be written \[\ \left[ {{x}^{3}}+{{\left( -\ 1 \right)}^{3}}+{{y}^{3}}-3\times x\times \left( -\ 1 \right)\times y \right]\]
So,
\[\Rightarrow {{x}^{3}}-1+{{y}^{3}}+3xy=\ \left[ {{x}^{3}}+{{\left( -\ 1 \right)}^{3}}+{{y}^{3}}-3\times x\times \left( -\ 1 \right)\times y \right]\].
\[\Rightarrow {{x}^{3}}-1+{{y}^{3}}+3xy=\left[ x+\left( -\ 1 \right)+y \right]\times \left[ {{x}^{2}}+{{\left( -\ 1 \right)}^{2}}+{{y}^{2}}-\left( x \right)\ \left( -\ 1 \right)-\left( -\ 1 \right)\ \left( y \right)-\left( y \right)\left( x \right) \right]\].
\[\Rightarrow {{x}^{3}}-1+{{y}^{3}}+3xy=\left( x-1+y \right)\left( {{x}^{2}}+1+{{y}^{2}}+x+y-xy \right)\].
Hence, the factors of \[{{x}^{3}}-1+ {{y}^{3}}+3xy\] are\[\left( x-1+y \right)\left( {{x}^{2}}+1+{{y}^{2}}+x+y-xy \right)\].
Therefore, the correct option for this question is (a).

Note:- Let us now learn about some other factorization identities. They are known as factorization identities such as \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] and \[\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab\]. Try not to make any calculation errors as if there are any errors in the starting of the solution, then the whole solution becomes more and more complex.