Let the transformed equation of $2x^4-8x^3+3x^2-1=0$ so that the term containing the cubic power of $x$ is absent be $2x^4+b{x}^2+cx+d=0$. Then $b=$

Given,

The transformed equation of $2x^4-8x^3+3x^2-1=0$ so that the term containing the cubic power of $x$ is absent be $2x^4+b{x}^2+cx+d=0$,

Now to find the transformed equation we will let $x=y+h$,

So the equation $2x^4-8x^3+3x^2-1=0$ becomes,

$2{\left(y+h\right)}^4-8{\left(y+h\right)}^3+3{\left(y+h\right)}^2-1=0$

$\Rightarrow 2\left(y^4+4y^{3}h+6y^2{h}^2+4y{h}^3+h^4\right)-8\left(y^3+3y^{2}h+3y{h}^2+h^3\right)+3{\left(y+h\right)}^2-1=0$

$\Rightarrow 2\left(y^4+6y^2{h}^2+4y{h}^3+h^4\right)+y^3\left(8h-8\right)-8\left(3y^{2}h+3y{h}^2+h^3\right)+3{\left(y+h\right)}^2-1=0$

Now eliminate $y^3$ we will make its coefficient zero,

So $8h-8=0\Rightarrow h=1$

So transfromed equation will be,

$\Rightarrow 2\left(y^4+6y^2+4y+1\right)-8\left(3y^2+3y+1\right)+3{\left(y+1\right)}^2-1=0$

$\Rightarrow 2y^4+12y^2+8y+2-24y^2-24y-8+3y^2+3+6y-1=0$$\Rightarrow 2y^4-9y^2-10y-4=0$

Now comparing with  $2x^4+b{x}^2+cx+d=0$ we get $b=-9$