Trivial ? Write the LHS as $$\begin{align*} x^2(2-x)^2&=\Big[1-(1-x)\Big]^2\Big[1+(1-x)\Big]^2\\ &=\Big[1-(1-x)^2\Big]^2\\ \end{align*}$$ Let, $~(1-x)^2=t$. Then, the whole equation transforms as $$(1-t)^2=1+2t~\Leftrightarrow~t(t-4)=0$$ which implies that $t=0,4$. Solving from here, we get $$\boxed{~x=~(-1)~,~1~,~3~}$$

We expand $x^2(2-x)^2$ this gives us $4x^2-4x^3+x^4$ and then we expand $1+2(1-x)^2$, which is $2x^2-4x+3$. We set them equal to each other, and then subtract $3$ from both sides leaving us with $4x^2-4x^3+x^4-3=2x^2-4x+3-3$. We simplify giving us, $4x^2-4x^3+x^4-3=2x^2-4x$. We then add $4x$ to both sides giving us, $4x^2-4x^3+x^4-3+4x=2x^2-4x+4x$. We simplify giving us, $4x^2-4x^3+x^4-3+4x=2x^2$. We subtract $2x^2$ from both sides, and we simplify giving us $4x^2-4x^3+x^4-3+4x-2x^2=2x^2-2x^2$. We set equal to $0$. So now we have $x^4-4x^3+2x^2+4x-3=0$. We factor giving us, $(x-1)^2(x+1)(x-3)=0$. By zero product property the solutions are, $x=1$, $x=-1$, and $x=3$. EDIT: Compared to the solution above, mine was a little bashy.

why you uguys snipe me like that

Actually, I wanted to submit the questions in Bangladesh Contests(2018). Can anyone please tell me how can I do that?

Here you go. https://artofproblemsolving.com/community/c6h1071990p4665221 Follow the instructions accordingly.

SINAN_EXPERT wrote: Actually, I wanted to submit the questions in Bangladesh Contests(2018). Can anyone please tell me how can I do that? I have done the work for you.Posted the links of the BdMO 2018 Problems to add them at Bangladesh Contest Section.

SINAN_EXPERT wrote: Solve: $x^2(2-x)^2=1+2(1-x)^2$ Where $x$ is real number. Solution. Noticing that $x^2(2-x)^2=\left(2x-x^2\right)^2=\left((1-x)^2-1\right)^2=(1-x)^4-2(1-x)^2+1$, we obtain that $$\begin{align*}x^2(2-x)^2=1+2(1-x)^2\iff&(1-x)^4-2(1-x)^2=2(1-x)^2\\ \iff&(1-x)^2\left[(1-x)^2-4\right]=0\\ \iff&(1-x)^2(x+1)(x-3)=0, \end{align*}$$ which yields the solution set $\{1,-1,3\}$. $\blacksquare$

integrated_JRC wrote: Here you go. https://artofproblemsolving.com/community/c6h1071990p4665221 Follow the instructions accordingly. Nope, it's not working.

SINAN_EXPERT wrote: Solve: $x^2(2-x)^2=1+2(1-x)^2$ Where $x$ is real number. Solution: $x^2(2-x)^{2} \\= (2x-x^2)^{2} \\= (x^2-2x)^{2}$ observe that , $2(1-x)^2 \\= 2(x^2-2x+1) \\= 2x^2-4x+2 \\= (2x-2)(x-1)$ we obtain that $(x^2-2x)^{2} = 1+(2x-2)(x-1)\\ \iff (x^2-2x)^{2}-1 = (2x-2)(x-1) \\\iff (x^2-2x+1)(x^2-2x-1)= (x-1)2(x-1) \\\iff (x-1)^{2}(x^2-2x-1) = 2(x-1)^{2} \\\iff (x-1)^{2}(x^2-2x-1)- 2(x-1)^{2} = 0 \\\iff (x-1)^{2}(x^2-2x-3)=0 \\\iff (x-1)^2(x+1)(x-3)=0$ so $x \in \{-1,1,3\}$

integrated_JRC wrote: Trivial ? Write the LHS as $$\begin{align*} x^2(2-x)^2&=\Big[1-(1-x)\Big]^2\Big[1+(1-x)\Big]^2\\ &=\Big[1-(1-x)^2\Big]^2\\ \end{align*}$$ Let, $~(1-x)^2=t$. Then, the whole equation transforms as $$(1-t)^2=1+2t~\Leftrightarrow~t(t-4)=0$$ which implies that $t=0,4$. Solving from here, we get $$\boxed{~x=~(-1)~,~1~,~3~}$$ I feel the difference of squares part is unnecessary? Note that: $$x^2(2 - x)^2 = (x(2 - x))^2 = (2x - x^2)^2$$ We see that $2x - x^2 = 1 - (1 - x)^2$. Let's replace that into the equation. $$(1 - (1 - x)^2)^2 = 1 + 2(1 - x)^2$$ Replace $(1 - x)^2$ as $y$. $$(1 - y)^2 = 1 + 2y$$ $$y^2 - 2y + 1 = 1 + 2y$$ Rearranging yields: $$y^2 - 4y = 0$$ Hence $y = 4, 0$. This implies that $x$ can have values $1, -1, 3$. Done

Solve for x over the real numbers: x^2 (2 - x)^2 = 2 (1 - x)^2 + 1 Expand out terms of the left hand side: x^4 - 4 x^3 + 4 x^2 = 2 (1 - x)^2 + 1 Expand out terms of the right hand side: x^4 - 4 x^3 + 4 x^2 = 2 x^2 - 4 x + 3 Subtract 2 x^2 - 4 x + 3 from both sides: x^4 - 4 x^3 + 2 x^2 + 4 x - 3 = 0 The left hand side factors into a product with three terms: (x - 3) (x - 1)^2 (x + 1) = 0 Split into three equations: x - 3 = 0 or (x - 1)^2 = 0 or x + 1 = 0 Add 3 to both sides: x = 3 or (x - 1)^2 = 0 or x + 1 = 0 Take the square root of both sides: x = 3 or x - 1 = 0 or x + 1 = 0 Add 1 to both sides: x = 3 or x = 1 or x + 1 = 0 Subtract 1 from both sides: Answer: | | x = 3 or x = 1 or x = -1

Olympus_mountaineer wrote: SINAN_EXPERT wrote: Actually, I wanted to submit the questions in Bangladesh Contests(2018). Can anyone please tell me how can I do that? I have done the work for you.Posted the links of the BdMO 2018 Problems to add them at Bangladesh Contest Section. I'd like to add BdMO 2020 problem to the section. Any suggestions on How would I do that?