Calculator is for the dull

Let $9998=n$ then the expression turns to:

$(n-1)^2-3n^2+3(n+1)^2$

$= n^2-2n+1-3n^2+3n^2+6n+3$

$=n^2+4n+4$

$=(n+2)^2$

Re-substituting $9998=n$ we have:

$(n+2)^2=(9998+2)^2=(10000)^2=\boxed{100000000}$

$\Rightarrow \color{#D61F06}{9997^2}-\color{#20A900}{3}×\color{#3D99F6}{9998^2}+\color{#20A900}{3}×\color{#EC7300}{9999^2}$

$\Rightarrow \color{#D61F06}{9997^2}-\color{#20A900}{3}[\color{#3D99F6}{9998^2}-\color{#EC7300}{9999^2}]$

$\Rightarrow \color{#D61F06}{9997^2}-\color{#20A900}{3}[\color{#BA33D6}{199997}×(\color{grey}{-1})]$

$\Rightarrow \color{#D61F06}{9997^2}+\color{#BA33D6}{199997}×\color{#20A900}{3}$

$\Rightarrow (\color{magenta}{10000}-\color{#20A900}{3})^2+[(\color{magenta}{10000}+\color{#D61F06}{9997})×\color{#20A900}{3}]$

Let $\color{magenta}{10000}=a.$

$\Rightarrow (a-\color{#20A900}{3})^2+(a+\color{#D61F06}{9997})×\color{#20A900}{3}$

$\Rightarrow a^2-\color{#20A900}{3}a+\color{#E81990}{30000}$

$\Rightarrow \color{magenta}{10000^2}-\color{#E81990}{30000}+\color{#E81990}{30000}=\boxed{\color{#624F41}{100000000}}$

Let $9997=x$

$x^2-3(x +1)^2+3(x+2)^2$

$x^2-3(x^2+2x+1)+3(x^2+4x+4)$

$x^2-3x^2-6x-3+3x^2+12x+12$

$x^2+6x+9$

$(x+3)^2$

Substitute $x$

$(9997+3)^2$

$10000^2$

$100000000$

Let $a=10^4$ . Then we have:

$\begin{aligned} (a-3)^2 - 3(a-2)^2 + 3(a-1)^2 & = a^2 - 6a + 9 - 3(a^2 - 4a + 4) + 3(a^2 - 2a+1) \\ & = a^2 = 10^8 = \boxed{\text{100 000 000}} \end{aligned}$

x= 9999 (9997)^2 - 3 (9998)^2 + 3 (9999)^2 (x-2)^2 - 3 (x - 1)^2 + 3 (x)^2 x^2 -4x +4 - 3x^2 +6x -3 +3x^2 x^2 +2x +1 (x+1)^2 (9999 +1 )^2 (10000)^2 = 100000000

nihar mahajan u r genius