Mathematical Proof - Cube Roots of Perfect Cubes Using Vedic Mathematics 
Please refer this site. A method is provided for finding cube
  roots of perfect cubes.
As per the method explained, suppose we are finding cube root of
  $157464$
First we write as  $157,\quad 464$
Last digit of $464$ is $4.$ Hence RHS=$4$
$157-5^3 \ge 0$ ($5$ is the maximum). So LHS$=5$
Hence, $\sqrt[3]{157464}=54$

How we can prove this mathematically? Please give directions on how to start towards writing a proof for this. Thanks.
 A: Lemma. Then function $x\mapsto x^3:\mathbb Z\to\mathbb Z$ induces a bijection $\mathbb Z/10\mathbb Z\to \mathbb Z/10\mathbb Z$.
Proof. It's enough to show that $x^3\equiv y^3\pmod{10}$ implies $x\equiv y\pmod{10}$.
If $x^3\equiv y^3\pmod{10}$, then $x\equiv x^3\equiv y^3\equiv y\pmod{2}
$ and $xy^2\equiv x^5y^2\equiv (xy)^2x^3\equiv (xy)^2y^3\equiv x^2y^5\equiv x^2y\pmod{5}$, hence $xy(x-y)\equiv 0\pmod{5}$.
Assume $x-y\not\equiv 0\pmod{5}$ so that (wlog) $x\equiv 0\pmod{5}$.
From $0\equiv x^3-y^3=(x-y)(x^2+xy+y^2)\pmod{5}$  we get $0\equiv x^2+xy+y^2\equiv y^2\pmod{5}$ from which $y\equiv 0\equiv x\pmod{5}$ which contradicts $x\not\equiv y\pmod 5$.
Thus $x\equiv y\pmod 5$, hence $x\equiv y\pmod{10}$ which concludes the proof.
Proposition
Let $n\in\mathbb N$.
There exists unique $q,r\in\mathbb N$ with $r<10^3$ such that $n=10^3q+r$.
There exists unique $d,u\in\mathbb N$ with $u<10$ such that $r\equiv u^3\pmod{10}$ and $d^3\leq q<(d+1)^3$.
Proposition. Let $n=10^3q+r$ and $m=10d+u$ with $r<10^3$ and $u<10$.
If $n=m^3$ then $r\equiv u^3\pmod{10}$ and $d^3\leq q<(d+1)^3$.
Proof. Clearly, we have $n\equiv r\equiv u^3\pmod{10}$.
If $q<d^3$, then $q\leq d^3-1$ hence $10^3d^3\leq n=10^3q+r\leq 10^3d^3+r-10^3$ from which $10^3\leq r$ - a contradiction which proves $d^3\leq q$.
If $q\geq (d+1)^3$, then $(10d+u)^3\geq (10d+10)^3$ from which $u\geq 10$ - a contradiction which proves $q<(d+1)^3$.
