Let's do some calculus! (33)

$\begin{aligned} (x+y)^4 & = x-y & \small \color{#3D99F6}{\text{Differentiate both sides wrt }x} \\ 4(x+y)^3 \left(1+\frac {dy}{dx} \right) & = 1- \frac {dy}{dx} \ \ ...(0) & \small \color{#3D99F6}{\text{Put }\frac {dy}{dx}=0} \\ 4(x+y)^3 & = 1 \\ (x+y)^3 & = \frac 14 \ \ ...\color{#3D99F6}{(1)} \\ x & = \frac 1{\sqrt[3]4} - y \ \ ...\color{#D61F06}{(2)} \end{aligned}$

Now consider $\dfrac {d^2y}{dx^2}$ by differentiating equation $(0)$ .

$\begin{aligned} 4(x+y)^3 \left(1+\frac {dy}{dx} \right) & = 1- \frac {dy}{dx} \ \ ...(0) \\ 12(x+y)^2 \left(1+\frac {dy}{dx} \right)^2 + 4(x+y)^3 \left(\frac {d^2y}{dx^2} \right) & = - \frac {d^2y}{dx^2} & \small \color{#3D99F6}{\text{Substituting }\frac {dy}{dx} = 0, \ (x+y)^3 = \frac 14} \\ 12 \left(\frac 14 \right)^\frac 23 + 4\cdot \frac 14 \left(\frac {d^2y}{dx^2} \right) & = - \frac {d^2y}{dx^2} \\ \implies \frac {d^2y}{dx^2} & = - 6 \left(\frac 14 \right)^\frac 23 < 0 \end{aligned}$

Therefore, $y$ is maximum when $\dfrac {dy}{dx} = 0$ and $(x+y)^3 = \dfrac 14$ , and that:

$\begin{aligned} \color{#3D99F6}{(x+y)^4} & = \color{#D61F06}{x}-y \\ \color{#3D99F6}{\left(\frac 14 \right)^\frac 43_{(1)}} & = \color{#D61F06}{\left(\frac 14 \right)^\frac 13 - y_{(2)}} - y \\ \implies y & = \frac 12 \left(\left(\frac 14 \right)^\frac 13 - \left(\frac 14 \right)^\frac 43 \right) \\ & = \frac 12 \cdot \frac 1{\sqrt[3]4} \left(1-\frac 14 \right) \\ & = \frac 12 \cdot \frac {\sqrt[3]2}{\sqrt[3]4 \cdot \sqrt[3]2} \left(\frac 34 \right) \\ \implies y_{max} & = \frac {3\sqrt[3]2}{16} \end{aligned}$

$\implies a+b+c = 3+2+16 = \boxed{21}$

As in the maximum y case its positive so is x.better substitute $x=y(secz)^2$ Now the rest is quite easy as it yields $tan z=√(2/3)$

${(x+y)}^4 = x-y \quad\quad \Rightarrow y = {\dfrac12}\times{\left((x+y)-{(x+y)}^4\right)}$

Let $(x+y)=t \quad\quad \Rightarrow y = {\dfrac12}\times{\left(t-t^4\right)}$

$\dfrac{dy}{dt}=0$ at $t = \dfrac{1}{\sqrt[3]{4}} , \qquad \dfrac{d^{2}y}{d{t}^2} >0$

$y_{max} = \dfrac{3\cdot{\sqrt[3]{2}}}{16}$

(x+y)^{4} = x - y

Taking derivative of both sides using implicit differentiation:

4 (x + y)^{3} [1 + y'] = 1 -y'

Putting y' on same side we have

4y' (x +y)^{3} + y' = 1 - 4 (x + y)^{3}

Solving for y'

y' = 1 - 4 (x +y)^{3}/((4(x +y)^{3} + 1))

Since we're solving for a maximum, we set this to zero, and after some very painless algebra you get

(x + y)^{3} = 1/4, solve for x, x = (1/4)^{1/3} - y

Replace into the given equation for x

((1/4)^{1/3} - y) +y)^{4} = (1/4)^{1/3} - y -y

Simplifying you get the following ((1/4)^{1/3})^{4} = (1/4)^{1/3} - 2y

Solving for y

2y = ( (1/4)^{1/3} (1/4)^{1/3})^{4}

Thus y = (after simplification) 3 (2)^{1/3}/16

3 + 2 + 16 = 21

Draw the graph of the function.. it shows the graph of Y = 2^(3/2) X^4 transforming axes like y=-x -> X and y=x -> Y from original function. Clearly, it attains the maximum value at some positive x when slope of the graph w.r.t X and Y becomes 45° at some x. At that point, X = 2^(-2/3) and Y = 2^(-8/3) ... Hardly any calculus required!! THERE YOU GO !!