Addend as a Factor of a Cube

Given that $\dfrac {a+b}a = c^3$ , $\implies 1 + \dfrac ba = c^3$ . Since the smallest $\dfrac ba = \dfrac 11 = 1$ and the largest $\dfrac ba = \dfrac {99}1 = 99$ , then $2 < c^3 < 100 \implies 2 \le c \le 4$ .

From $1 + \dfrac ba = c^3 \implies b = a(c^3 -1)$ . Then $a+b+c = a + a(c^3 -1) + c = ac^3 + c$ . This means that for a particular value of $c$ , the largest $a+b+c$ is the one with the largest $a$ . In formula, we have $\max(a+b+c) = a_{\max} c^3 + c$ . And $a_{\max} = \left \lfloor \dfrac {99}{c^3-1} \right \rfloor$ , where $\lfloor \cdot \rfloor$ denotes the floor function .

Then we have:

$\begin{array} {lll} c = 2 & \implies a_{\max} = \left \lfloor \dfrac {99}{2^3-1} \right \rfloor = 14 & \implies \max(a+b+c) = 14(8) + 2 = 114 \\ \phantom0 \\ c = 3 & \implies a_{\max} = \left \lfloor \dfrac {99}{3^3-1} \right \rfloor = 3 & \implies \max(a+b+c) = 3(27) + 3 = 84 \\ \phantom0 \\ c = 4 & \implies a_{\max} = \left \lfloor \dfrac {99}{4^3-1} \right \rfloor = 1 & \implies \max(a+b+c) = 1(64) + 4 = 68 \end{array}$

Therefore the largest value of $a+b+c = \boxed{114}$ .

The fraction $\frac{a+b}{a}$ can be rewritten as $1+\frac{b}{a}$ . Meaning, $\frac{b}{a}$ is equal to $c^{3} - 1$ . The highest possible value for $\frac{b}{a}$ if both variables are positive integers less than $100$ is $\frac{99}{1}$ . Thus we must find cubes less than $100$ and subtract them by $1$ .

$\begin{array}{lcrrr} 1^{3} & = & 1 & \implies & 0 \\[0.25em] 2^{3} & = & 8 & \implies & 7\\[0.25em] 3^{3} & = & 27 & \implies & 26 \\[0.25em] 4^{3} & = & 64 & \implies & 63 \end{array}$

If $c^{3} - 1$ is $0$ , the solutions will be $\frac{0}{n} = 0$ (with $n$ being any number satisfying the mentioned conditions). But $b > 0$ , thus the cube cannot be $1$ .

We then need to list the divisors and dividends of the quotient $c^{3} - 1$ .

$\begin{array}{rcl} 7 & \implies & \dfrac71 , \dfrac{14}{2}, \dfrac{21}{3}, \ldots , \dfrac{91}{13}, \dfrac{98}{14} \\[1.25em] 27 & \implies & \dfrac{27}{1}, \dfrac{54}{2}, \dfrac{81}{3} \\[1.25em] 63 & \implies & \dfrac{63}{1} \end{array}$

We see that the largest pair of divisor and dividend is $\frac{98}{14}$ .

$\begin{array}{ll} \begin{aligned} \dfrac{a+b}{a} = & \space c^{3} \\[0.5em] \dfrac{14+98}{14} = & \space 2^{3} \end{aligned} & \implies a + b + c = 14 + 98 + 2 = \boxed{114} \end{array}$

Given a, b, c are positive integers $0 < a, b < 100$ ,

$\cfrac{a+b}{a} = c^3 \Rightarrow 1+ \cfrac{b}{a} = c^3 \Rightarrow b = na$ for some postive integer $\min\left( \cfrac{b}{a} \right) = 1 \le n \le \max\left( \cfrac{b}{a} \right) = 99 \\ \cfrac{(1+n)a}{a} = c^3 \Rightarrow c^3 \leq 100 \Rightarrow c = 2, 3, 4$

As $b=na < 100, c \in [2,4], c\downarrow \Rightarrow n\downarrow \Rightarrow a\uparrow \Rightarrow b\uparrow=na$ ,

To maximize $a+b+c$ we choose smallest c

$c = 2, n=7, b=na=7a < 100 \Rightarrow$ $\begin{aligned} \max(a) &= 14 \\ \max(b) &= 98 \\ \max( a+b+c ) &= 14+98+2 = \boxed{114} \end{aligned}$