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

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

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

\(-1 = x^6-x^3+x\)

\(-1 = x^2+x\)

\(x^2+x+1 = 0\)

\(B^2 - 4 *C = 1^2 - 4 *1 *1 = -3\)

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 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?