The roots of the equation ${\left(3-x\right)}^4+{\left(2-x\right)}^4={\left(5-2x\right)}^4$  are

The given equation is ${\left(3-x\right)}^4+{\left(2-x\right)}^4={\left(5-2x\right)}^4$         ……(i)

Put $3-x=a$  and $2-x=b$

Eq. (i) be comes $a^4+b^4={\left(a+b\right)}^4$

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

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

$\therefore$   $a=0\Rightarrow 3-x=0\Rightarrow x=3$

      $b=0\Rightarrow 2-x=0\Rightarrow x=2$

and  $2a^2+3ab+2b^2=0$

        $2{\left(\frac{a}{b}\right)}^2+3\left(\frac{a}{b}\right)+2=0$

       $\frac{a}{b}=\frac{-3\pm \sqrt{9-16}}{4}$

      $\frac{3-x}{2-x}=\frac{-3\pm i\sqrt{7}}{4}$

$\Rightarrow$  $\frac{3-x}{2-x}-1=\frac{-7\pm i\sqrt{7}}{4}$

$\Rightarrow$  $\frac{1}{2-x}=\frac{-7\pm i\sqrt{7}}{4}$

    $2-x=\frac{4}{-7\pm i\sqrt{7}}$

    $x-2=\frac{4}{-7\pm i\sqrt{7}}$

        $x=\frac{-18\pm 2i\sqrt{7}}{-7\pm i\sqrt{7}}$

Taking ‘+’ sign

       $x=\frac{\left(-18+2i\sqrt{7}\right)\left(-7-i\sqrt{7}\right)}{\left(-7+i\sqrt{7}\right)\left(-7-i\sqrt{7}\right)}$

          $=\frac{140+4i\sqrt{7}}{56}$

$\therefore$   $x=\frac{35+i\sqrt{7}}{14}$

Taking ‘-‘ sign,    $x=\frac{35-i\sqrt{7}}{14}$

Hence, roots of Eq. (i) are $2,3,\frac{35\pm i\sqrt{7}}{14}$

i.e.,   Two real and two imaginary.