A Ball in a Cone? Is that something new?

$\begin{aligned} R+h & = & 112 \quad \cdots (1) \\ h+l & = & 144 \quad \cdots (2) \\ l+R & = & 128 \quad \cdots (3) \end{aligned}$

$(1)+(2)+(3)$ gives,

$2 \left( l+R+h \right) = 384 \implies \left( l+R+h \right) = 192 \quad \cdots (4)$

$(4) - (1)$ gives,

$\color{#D61F06}{l=80}$

$(4) - (2)$ gives,

$\color{#20A900}{R=48}$

$(4) - (3)$ gives,

$\color{#302B94}{h=64}$

Consider the diagram above of half-conical section where $\theta$ is exactly half of the original angle made at the apex.

Using trigonometry, we have,

$\cos \theta = \dfrac{\color{#302B94}{h}}{\color{#D61F06}{l}} = \dfrac{64}{80} = \dfrac{4}{5} \\ \therefore \sin \theta = \sqrt{1- \cos^2 \theta} = \dfrac{3}{5}$

From the figure, we observe that,

$\sin \theta = \dfrac{r}{\color{#302B94}{h}-r} \\ \therefore \dfrac{r}{\color{#302B94}{h}-r} = \dfrac 35 \implies \dfrac{r}{64-r} = \dfrac 35 \implies \color{maroon}{r = 24}$

Thus,

$\begin{aligned} \dfrac{\text{Volume of ball}}{\text{Volume of cone}} &=& \dfrac{\frac {4}{\cancel{3}} \cancel{\pi} r^3}{\frac{1}{\cancel{3}} \cancel{\pi} R^2 h} \\ \\ &=& \dfrac{\cancel{4}^1 \cdot 24^3}{48^2 \cdot \cancel{64}^{16}} \\ \\ &=& \dfrac{\cancel{24^3}^{24}}{\cancel{24^2}^1 \cdot 2^2 \cdot 16} \\ \\ &=& \dfrac{24}{64} \\ \\ &=& \dfrac 38 = \dfrac{\color{#D61F06}{x}}{\color{#3D99F6}{y}} \end{aligned}$

Hence,

$\begin{aligned} \left( x^2+3xy+y^2 \right) \left(x^2+xy+y^2 \right) &= \left[ {\left(x+y\right)}^2 + xy \right] \left[ {\left(x+y\right)}^2 - xy \right] \\ &= {\left(\color{#D61F06}{x}+\color{#3D99F6}{y}\right)}^4 - {(\color{#D61F06}{x} \color{#3D99F6}{y})}^2 \\ & = {(3+8)}^4 - {(3 \times 8)}^2 \\ & = 11^4 - 24^2 \\ & = 14641-576 \\ &= \boxed{14065} \end{aligned}$

$R+h =112$ $..........(1)$

$h+l=144$ $..........(2)$

$l+R=128$ $..........(3)$

$(1) + (2) + (3)$

$2(R+h+l) = 384$

$R+h+l=192$ $..........(4)$

$(4) - (2)$

$R=48$ ; By Triple Pythagoras, $h = 64 ; l=80$

$r = \frac{2Rh\times \frac{1}{2}}{\frac{1}{2}\times(2R+2l)}$

$r= \frac{2Rh}{2(R+l)}$

$r= \frac{Rh}{R+l}$

$r = \frac{3072}{128}$

$r=24$

Volume of Ball $:$ Volume of Cone

$\frac{4}{3} \times \pi \times r³ : \frac{1}{3} \times \pi \times R² \times h$

$4 \times 24 \times 24 \times 24 : 48 \times 48 \times 64$

$3:8 = x:y$

$x= 3 ; y =8$

Note that $x² + 3xy + y² = x² + 2xy + y² + xy = (x+y)² + xy$ and

$x² + xy + y² = x² + 2xy + y² - xy = (x+y)²-xy$

Now,

$((x+y)² +xy )( (x+y)² -xy) = (x+y)^4 -((xy)^2)$

$11^4 - 24^2 = 14641 -576$

$= \boxed{14065}$