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Math Revolution GMAT Instructor V
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What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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[Math Revolution GMAT math practice question]

What is the number of roots of the equation $$\frac{(x^3-1)}{(x^2+x+1)}=x$$?

$$A. 1$$
$$B. 2$$
$$C. 3$$
$$D. 4$$
$$E. no solution$$

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Originally posted by MathRevolution on 22 Oct 2018, 01:29.
Last edited by MathRevolution on 25 Oct 2018, 23:27, edited 1 time in total.
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the number of roots of the equation $$\frac{(x^3-1)}{(x^2+x+1)}=x$$?

$$A. 0$$
$$B. 1$$
$$C. 2$$
$$D. 3$$
$$E. no solution$$

You know, $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
Factorizing numerator of the given expression(LHS), we have
$$\frac{(x^3-1)}{(x^2+x+1)}=x$$
Or, $$\frac{(x-1)(x^2+x+1)}{(x^2+x+1)}=x$$
Or, x-1=x
Substracting x from both LHS & RHS,
-1=0
Therefore, No Solution.
Ans. (E)

Another approach:-
$$\frac{(x^3-1)}{(x^2+x+1)}=x$$
Or, $$(x^3-1)=x(x^2+x+1)$$
Or, $$x^3-1=x^3+x^2+x$$
Or, $$x^3+x^2+x=x^3-1$$
Or, $$x^2+x+1=0$$ is a quadratic equation with a=b=c=1
$$D=b^2-4ac=1^2-4*1*1=-3<0$$; so no real roots.

Hence, No Solution exists.

Ans. (E)
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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Ans is E as after solving we will get x2+x+1=0 and D is negative for this equation. Hence No real solution is possible.
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the number of roots of the equation $$\frac{(x^3-1)}{(x^2+x+1)}=x$$?

$$A. 0$$
$$B. 1$$
$$C. 2$$
$$D. 3$$
$$E. no solution$$

$$\frac{(x^3-1)}{(x^2+x+1)}=x$$

$$x^3-1 = x(x^2+x+1)$$

$$x^3-1 = x^3+x^2+x$$

$$-1 = x^6-x^3+x$$

$$-1 = x^2+x$$

$$x^2+x+1 = 0$$

$$B^2 - 4 *C = 1^2 - 4 *1 *1 = -3$$

since the root is less than 0, there are no solutions

pushpitkc, interesting if I couldn't use a factoring method, like product of C equals the sum of B Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82
Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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=>

The original condition $$\frac{(x^3-1)}{(x^2+x+1)}=x$$ is equivalent to $$x – 1 = x$$ as shown below:

$$\frac{(x^3-1)}{(x^2+x+1)}=x$$
$$=> \frac{(x-1) (x^2+x+1)}{(x^2+x+1)} = x$$ by factoring
$$=> x – 1 = x$$

But $$x – 1 = x$$ has no solution.

Therefore, the answer is E.
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MathRevolution wrote:
=>

The original condition $$\frac{(x^3-1)}{(x^2+x+1)}=x$$ is equivalent to $$x – 1 = x$$ as shown below:

$$\frac{(x^3-1)}{(x^2+x+1)}=x$$
$$=> \frac{(x-1) (x^2+x+1)}{(x^2+x+1)} = x$$ by factoring
$$=> x – 1 = x$$

But $$x – 1 = x$$ has no solution.

Therefore, the answer is E.

math revoultion
I completely agree that there are no solutions for the equation. Bbt Answer option A tells us the number of solutions is 0, which is the same as No solution for any equation. Only if an equation has 0 solutions, will it have no equations.
Where am I wrong here?
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8005
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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nitesh50 wrote:
MathRevolution wrote:
=>

The original condition $$\frac{(x^3-1)}{(x^2+x+1)}=x$$ is equivalent to $$x – 1 = x$$ as shown below:

$$\frac{(x^3-1)}{(x^2+x+1)}=x$$
$$=> \frac{(x-1) (x^2+x+1)}{(x^2+x+1)} = x$$ by factoring
$$=> x – 1 = x$$

But $$x – 1 = x$$ has no solution.

Therefore, the answer is E.

math revoultion
I completely agree that there are no solutions for the equation. Bbt Answer option A tells us the number of solutions is 0, which is the same as No solution for any equation. Only if an equation has 0 solutions, will it have no equations.
Where am I wrong here?

You are right.
0 should be eliminated from the choices.
_________________ Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?   [#permalink] 25 Oct 2018, 23:27
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