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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PKN
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73% (00:56) correct 
