Question

The factors of ${x^3} - 1 + {y^3} + 3xy$ are:
A. $\left( {x - 1 + y} \right)\left( {{x^2} + 1 + {y^2} + x + y - xy} \right)$
B. $\left( {x + y + 1} \right)\left( {{x^2} + {y^2} + 1 - xy - x - y} \right)$
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: Here we need to use the formula where we can compare the given term by the formula:
${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$
So we can write ${x^3} - 1 + {y^3} + 3xy$ in this form and then get the required answer.

Complete step by step solution:
Here we are given that we need to find the factors of ${x^3} - 1 + {y^3} + 3xy$ which means we need to write it in the form of the multiplication of the two terms. So we need to see which formula is to be used.
We know that as we have the formula where we can get:
${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$
Now we can compare this formula with the given equation which is ${x^3} - 1 + {y^3} + 3xy$
Now we can write this equation as:
${x^3} + {\left( { - 1} \right)^3} + {y^3} - 3\left( { - 1} \right)xy$
If we compare this with ${a^3} + {b^3} + {c^3} - 3abc$
Then we can say that:
$
  a = x \\
  b = - 1 \\
  c = y \\
 $
Now we can simply substitute the values of all the variables of the formula with the given equation, then we will get:
 ${x^3} + {\left( { - 1} \right)^3} + {y^3} - 3\left( { - 1} \right)xy$$ = \left( {x - 1 + y} \right)\left( {{x^2} + {{\left( { - 1} \right)}^2} + {y^2} - x\left( { - 1} \right) - \left( { - 1} \right)y - xy} \right)$
${x^3} + {\left( { - 1} \right)^3} + {y^3} - 3\left( { - 1} \right)xy$$ = \left( {x - 1 + y} \right)\left( {{x^2} + 1 + {y^2} + x + y - xy} \right)$
Hence whenever we are given the equation and the factors are to be found then we simply need to apply the formula and then compare the terms and get the factors in the simplified form.
So we have got that:
${x^3} + {\left( { - 1} \right)^3} + {y^3} - 3\left( { - 1} \right)xy$$ = \left( {x - 1 + y} \right)\left( {{x^2} + 1 + {y^2} + x + y - xy} \right)$

Hence we can say that A) is the correct option out of the given four options.

Note:
Here the student must know the general formula of all the cubic as well as the square option. If we are given to find the factors of $\left( {{a^2} + {b^2} + 2ab - {c^2}} \right)$ then we can write it as $\left( {{{\left( {a + b} \right)}^2} - {c^2}} \right)$.
Now we can apply the formula ${x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)$ and get the factors in simplified form.