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Question: What loci are represented by the equation: \({{x}^{3}}-{{x}^{2}}-x+1=0\)?...

What loci are represented by the equation: x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0?

Explanation

Solution

Hint: We can start solving the above question by factorising the given equation i.e. x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0. The values of x obtained would give the loci of the equation.

Complete step-by-step solution -

We will find loci of the equation x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0 given in the question by simplifying it into factors which cannot be factorised further.
The given equation is x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0. Now, we will simplify the cubic equation by taking out the common terms.
In the above cubic equation, we will take x2{{x}^{2}} common from x3{{x}^{3}} and x2{{x}^{2}}. Thus, the above equation becomes
⇒x2(x−1)−x+1\Rightarrow {{x}^{2}}(x-1)-x+1
Now we will take -1 common from the constant term and x. Then the equation becomes ⇒x2(x−1)−1(x−1)\Rightarrow {{x}^{2}}(x-1)-1(x-1)
Now we will take common terms out from the above equation. So, the equation becomes (x−1)(x2−1)(x-1)({{x}^{2}}-1)
We know that a2−b2{{a}^{2}}-{{b}^{2}} can be expressed as (a+b)(a−b)(a+b)(a-b). Applying the same, we will express x2−1{{x}^{2}}-1 as (x+1)(x−1)(x+1)(x-1).
Now the equation x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0 can be expressed as (x+1)(x-1)(x-1)=0. The values of x can hence be obtained by equating each term to 0. Therefore, the possible values of x are x=-1, x=1 or x=1.
The values x=-1, x=1 and x=1 represent three straight lines. We can see that two lines are coincident i.e. x=1 and x=1 and these coincident lines are parallel to the line x=-1.
Hence, we get that the loci of the equation x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0 are straight lines.

Note: This problem can be solved in an alternate way. If all roots of the given cubic equation are real, then loci of that cubic equation represents straight lines. Let us check whether roots are real or not for the cubic equation x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0. We will find a root by trial and error method or by observing the equation. If we observe the equation, the sum of coefficients of all terms = 0. Hence, one root will be 1 which is real, that means x-1 is a factor of x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0. So, we will divide x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0 with x-1 we will get the quotient as x2−1{{x}^{2}}-1. Now the roots of x2−1{{x}^{2}}-1 are also the roots of x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0 . -1 and 1 are roots of x2−1{{x}^{2}}-1 . Here, -1 and 1 are also real roots. All roots are real. Hence the loci of x3−x2−x+1=0{{x}^{3}}-{{x}^{2}}-x+1=0 are straight lines.