Question

How do you simplify \[\dfrac{{{x^3} + 1}}{{x + 1}}\] ?

Answer 1

Hint: This is a very simple question to solve. \[{x^3} + 1\] can also be written as \[{x^3} + {1^3}\] .Here we will just expand the identity \[{x^3} + {y^3}\] and then divide as per question. Simplifying the expression is nothing but cancelling the like terms if observed and performing necessary operations. We know that \[\left( {{x^3} + {y^3}} \right) = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\] . Using this expansion of identity we will solve the given expression. Then we will use quadratic formula to solve the equation

Complete step by step solution:
Given that
 \[\dfrac{{{x^3} + 1}}{{x + 1}}\]
Now write the expanded form,
 \[\dfrac{{{x^3} + 1}}{{x + 1}} = \dfrac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)}}\]
Now we observe that the first bracket of numerator is exactly the same as the denominator. So we can cancel them easily.
 \[\dfrac{{{x^3} + 1}}{{x + 1}} = {x^2} - x + 1\]
Now if we observe that the remaining equation is a quadratic equation. So we will solve it by using a quadratic equation.
 \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Comparing this with the general quadratic equation \[a{x^2} + bx + c = 0\] we get \[a = 1,b = - 1\& c = 1\] .
Thus substituting the values,
 \[ = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}\]
On solving this,
 \[ = \dfrac{{1 \pm \sqrt {1 - 4} }}{2}\]
On solving the roots,
 \[ = \dfrac{{1 \pm \sqrt { - 3} }}{2}\]
Now we know that \[\sqrt { - 1} = i\] thus we get,
 \[ = \dfrac{{1 \pm \sqrt 3 i}}{2}\]
 So the roots are \[ \Rightarrow \dfrac{{1 + \sqrt 3 i}}{2} or \Rightarrow \dfrac{{1 - \sqrt 3 i}}{2}\]
These are our factors.
So, the correct answer is “ \[ \dfrac{{1 + \sqrt 3 i}}{2} or \dfrac{{1 - \sqrt 3 i}}{2}\] ”.

Note: Note that quadratic formula is used to find the roots of a given quadratic equation. Sometimes we can factorize the roots directly. But quadratic formulas can be used generally to find the roots of any quadratic equation. The value of discriminant is used to decide the type of roots so obtained such that roots are equal or different and are real or not.Also note that the root so obtained is imaginary but different.