$f\left(x\right)={\left(2x^2+1\right)}^4-{\left(2x^2-1\right)}^4={\left({\left(2x^2+1\right)}^2\right)}^2-{\left({\left(2x^2-1\right)}^2\right)}^2=$

$=\left[{\left(2x^2+1\right)}^2+{\left(2x^2-1\right)}^2\right]\cdot \left[{\left(2x^2+1\right)}^2-{\left(2x^2-1\right)}^2\right]=$

$=\left[4x^4+4x^2+1+4x^4-4x^2+1\right]\cdot \left[4x^4+4x^2+1-\left(4x^4-4x^2+1\right)\right]=$

$=\left(8x^4+2\right)\cdot \left[4x^4+4x^2+1-4x^4+4x^2-1\right]=$

$=\left(8x^4+2\right)\cdot 8x^2=64x^6+16x^2$

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$g\left(x\right)=4{\left(2\sqrt{3}{x}^2+1\right)}^2-8\sqrt{3}{x}^2=4\left(4\cdot 3\cdot x^4+4\sqrt{3}{x}^2+1\right)-8\sqrt{3}{x}^2=$

$=48x^4+16\sqrt{3}{x}^2+4-8\sqrt{3}{x}^2=48x^4+8\sqrt{3}{x}^2+4$   

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$h\left(x\right)={\left(2x^2+1\right)}^3-{\left(2x^2-1\right)}^3=$

$=\left(8x^6+3\cdot 4x^4\cdot 1+3\cdot 2x^2\cdot 1+1\right)-\left(8x^6-3\cdot 4x^4\cdot 1+3\cdot 2x^2\cdot 1-1\right)=$  

$=8x^6+12x^4+6x^2+1-8x^6+12x^4-6x^2+1=24x^4+2$

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$k\left(x\right)=4x^2{\left(4x^2+2\right)}^2-64x^4=4x^2\left(16x^4+16x^2+4\right)-64x^4=$

$=64x^6+64x^4+16x^2-64x^4=64x^6+16x^2$

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st(f)=st(k) oraz wielomiany f i k mają równe współczynniki przy odpowiednich potęgach zmiennej, więc wielomiany te są równe.

Odp. Wielomiany f i k są równe.