Originally Posted by

**grgrsanjay** the number of real values for x such that{$\displaystyle (x+1)^3$}$\displaystyle = x^3$ where {$\displaystyle x$} is the fractional part of x

i.e{2.635}=635

i think the answer is 1 cuz we get the answer right when we substitute 0 and i cannot think any other values that satisfy this equation

options are:

(A)0

(B)1

(C)4

(D)6

If $\displaystyle x^3$ is the fractional part of x then it must lie between 0 and 1, and therefore so must x.

Next, $\displaystyle (x+1)^3 = x^3 + 3x^2 + 3x + 1$. If the fractional part of this is $\displaystyle x^3$ then the remaining part $\displaystyle 3x^2+3x+1$ must be an integer. So we want to find numbers x between 0 and 1 that satisfy the equation $\displaystyle 3x^2+3x-n=0$ for some integer n. The solution to the quadratic equation is $\displaystyle x = \frac{-3+\sqrt{9+12n}}6$ (neglecting the negative square root because that leads to a negative value for x). That gives these solutions:

$\displaystyle \begin{array}{l|l}n&\multicolumn{1}{c}{x}\\\hline0 &0\\ 1&0.2637...\\2&0.4574...\\ 3&0.6180...\\ 4&0.7583...\\ 5&0.8844...\end{array}$

When n = 6 we get x = 1, which is too large. So there are six solutions in all.