What superpowers would you have?

All variables are assumed to represent integers. We want to solve $k^3 \; = \; 2(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4) \; = \; 2(a^2 + ab + b^2)^2$ Thus $k$ must be even, which implies that $(a^2+ab+b^2)^2$ is a multiple of $4$ , and hence $a^2+ab+b^2$ is even. This implies that both $a$ and $b$ are even, and hence $k = 2x$ , $a = 2y$ , $b = 2z$ where $x^3 \; = \; 4(y^2 + yz + z^2)^2$ This in turn implies that $x$ is even, and hence that $y^2+yz+z^2$ is even, and hence that $y$ and $z$ are both even. Thus $x = 2X$ , $y = 2Y$ , $z = 2Z$ where $X^3 \; = \; 8(Y^2 + YZ + Z^2)^2$ which implies that $X$ is even, so that $X = 2U$ where $U^3 \; = \; (Y^2 + YZ + Z^2)^2$ From this it is clear that $U = V^2$ and $Y^2 + YZ + Z^2 = V^3$ .

The smallest possible solution for $V$ of the equation $V^3 \; = \; Y^2 + YZ + Z^2$ with $Y,Z$ distinct positive integers is $V=7$ (with $Y=14$ and $Z=7$ ), and so the smallest possible value of $k=8U=8V^2$ is $8\times49=392$ .

The fact that $V=7$ can be established by showing that $4\times1^3,4\times2^3,\ldots,4\times6^3$ cannot be written as $(2Y+Z)^2 + 3Z^2$ for distinct positive integer $Y,Z$ . To do this, just run through the finite number of possible $Z$ values, checking that no $Y$ value is possible. For example, if $(2Y+Z)^2 + 3Z^2 = 4\times3^3 = 108$ , then $\begin{array}{rcl} Z = 1 & \Rightarrow & (2Y+1)^2 = 105 \\ Z=2 & \Rightarrow & (2Y+2)^2 = 96 \\ Z=3 & \Rightarrow & (2Y+3)^2 = 81 \; \Rightarrow \; Y = 3 = Z \\ Z=4 & \Rightarrow & (2Y+4)^2 = 60 \\ Z=5 & \Rightarrow & (2Y+5)^2 = 33 \\ Z\ge6 & \Rightarrow & (2Y+Z)^2 \le 0 \end{array}$

let $k^3 = a^4 + b^4 + (a+b)^4$ so:

$k^3 = 2(a^2 + ab + b^2)^2$

then $k$ need to be even so let $(1), k = 2n$ where $n$ is positive integer, by substituting we've:

$k^3 = 2(a^2 + ab + b^2)^2$

become

$(2n)^3 = 2^3n^3 = 2(a^2 + ab + b^2)^2$

divide both sides by $2$

$2^2n^3 = (a^2 + ab + b^2)^2$

let $(2), n = m^2$ where $m$ is positive integer then:

$2^2n^3 = 2^2(m^2)^3 = 2^2(m^3)^2 = (2m^3)^2 = (a^2 + ab + b^2)^2$

add root square to both sides (because $m, a, b$ is positive integers) so:

$2m^3 = a^2 + ab + b^2$

for $a^2 + ab + b^2$ can be even, $a,b$ must be even so: let $a = 2x, b=2y$ and we've $a,b$ are distinct positive integers, without loss of generality let $b = 2y = a + 2z = 2x + 2z = 2(x+z)$ then:

$2m^3 = a^2 + ab + b^2$

become

$2m^3 = (2x)^2 + (2x)(2(x+z)) + (2(x+z))^2$

divide the both sides by $2$ and factorize:

$m^3 = 2(3x^2 + 3xz + z^2)$ and let again $(3), m = 2l$ where $l$ is positive integer so:

$m^3 = 2(3x^2 + 3xz + z^2)$

$(2l)^3 = 2(3x^2 + 3xz + z^2)$

$l^3 = (3x^2 + 3xz + z^2)/4$

for $3x^2 + 3xz + z^2$ be multiple of $4$ must the both $x,z$ be even also unnecessary for $x,z$ be distinct positive integers, so when:

$x = z = 2, l^3 = (3x^2 + 3xz + z^2)/4 = 7 = 1\times1\times7$

$x = z = 4, l^3 = (3x^2 + 3xz + z^2)/4 = 28 = 2\times2\times7$

...

$x = z = 14, l^3 = (3x^2 + 3xz + z^2)/4 = 343 = 7\times7\times7$

we've $k = 2((2\times l)^2)$ from $(1), (2), (3)$ where $l = 7$ then $k = 392$

for those of us who think in code

public class LeastPositiveK {
public static void main(String[] args) {
    for (int k = 1; k < 1000; k++) {
        for (int a = 1; a <= 1000; a++) {
            for (int b = 1; b <= 1000; b++) {
                int kCube = k*k*k;
                int a4 = a*a*a*a;
                int b4 = b*b*b*b;
                int ab = a+b;
                int ab4 = ab*ab*ab*ab;
                int sumRight = a4+b4+ab4;
                if (sumRight == kCube && a != b) {
                    System.out.println("k="+k+", a="+a+", b="+b);
                }
            }
        }
    }
}
}