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# What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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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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Updated on: 25 Oct 2018, 23:27
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15% (low)

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80% (01:17) correct 20% (01:46) wrong based on 78 sessions

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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" 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. VP Status: Learning stage Joined: 01 Oct 2017 Posts: 1017 WE: Supply Chain Management (Energy and Utilities) Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x? [#permalink] ### Show Tags 22 Oct 2018, 02:24 1 1 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) _________________ Regards, PKN Rise above the storm, you will find the sunshine Intern Joined: 06 Feb 2017 Posts: 37 Location: India Schools: HBS '22, HEC '22 GMAT 1: 570 Q39 V28 GMAT 2: 620 Q49 V26 GPA: 4 Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x? [#permalink] ### Show Tags 22 Oct 2018, 04:31 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. _________________ I hope this helped. If this was indeed helpful, then you may say Thank You by giving a Kudos! VP Joined: 09 Mar 2016 Posts: 1230 Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x? [#permalink] ### Show Tags 22 Oct 2018, 08:09 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 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] ### Show Tags 24 Oct 2018, 01:18 => 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. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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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24 Oct 2018, 01:24
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.

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
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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25 Oct 2018, 23:27
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.

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.
_________________
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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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