Let A denote the set of all real numbers x su | vedclass

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(B) Given the equation: $x^3 - [x]^3 = (x - [x])^3$. Let ${x} = x - [x]$. Then the equation becomes $x^3 - [x]^3 = {x}^3$. We know that $x^3 - [x]^3 =

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$A$ contains an interval,but is not an interval

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(B) Given the equation: $x^3 - [x]^3 = (x - [x])^3$. Let ${x} = x - [x]$. Then the equation becomes $x^3 - [x]^3 = {x}^3$. We know that $x^3 - [x]^3 = (x - [x])(x^2 + x[x] + [x]^2)$. So,$(x - [x])(x^2 + x[x] + [x]^2) = (x - [x])^3$. This implies $(x - [x])[(x^2 + x[x] + [x]^2) - (x - [x])^2] = 0$. $(x - [x])[x^2 + x[x] + [x]^2 - (x^2 - 2x[x] + [x]^2)] = 0$. $(x - [x])[3x[x]] = 0$. This gives two cases: Case $1$: $x - [x] = 0 \implies x \in \mathbb{Z}$. Case $2$: $3x[x] = 0 \implies x = 0$ or $[x] = 0$. If $[x] = 0$,then $0 \le x Combining these,$A = \mathbb{Z} \cup [0, 1)$. Since $A$ contains the interval $[0, 1)$ but also includes isolated points like $\dots, -2, -1, 2, 3, \dots$,it is not an interval. Thus,$A$ contains an interval,but is not an interval.

Let $A$ denote the set of all real numbers $x$ such that $x^3-[x]^3=(x-[x])^3$,where $[x]$ is the greatest integer less than or equal to $x$. Then,

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