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MathRevolution
[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 :?
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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.
Answer: E
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MathRevolution
=>

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


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?
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nitesh50
MathRevolution
=>

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


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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