The equation ${(3-x)}^4+{(2-x)}^4={(5-2x)}^4$  has

The given equation is ${(3-x)}^4+{(2-x)}^4={(5-2x)}^4$ 

Let $a={(3-x)}^{ },b={(2-x)}^{ }$ $\Rightarrow a^4+b^4={(a+b)}^4$

$\Rightarrow a^4+b^4=(a^2+2ab+b^2)(a^2+2ab+b^2)$

$\Rightarrow a^4+b^4=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$

$\Rightarrow 4a^{2}b+6a^2{b}^2+4a{b}^3=0$

$\Rightarrow 2ab(2a^2+3ab+2b^2)=0$

$\Rightarrow a=0$  (or)  $b=0$  (or)  $2a^2+3ab+2b^2=0$

$\Rightarrow {\left(3-x\right)}^4=0$ (or) $\Rightarrow {\left(2-x\right)}^4=0$ (or)$2{\left(3-x\right)}^2+3\left(3-x\right)\left(2-x\right)+2{\left(2-x\right)}^2=0$

$\Rightarrow x=3$  (or) $x=2$ (or)  $7x^2-35x+44=0$

Since  $7x^2-35x+44=0$  has non-real roots,

therefore, the given equation have only two real roots.