$ \begin{array}{l}{{x}^{4}}-2{{x}^{2}}+k\\={{x}^{4}}-2{{x}^{2}}+1+k-1\\={{({{x}^{2}}-1)}^{2}}+k-1\\If\,k=1\,\text{then}\,\text{the}\,\text{term}\,\text{will}\,\text{be}\,\text{a}\,\text{perfect}\,\text{square}\text{.}\end{array}$

$ \displaystyle \begin{array}{l}p-2q=4\\Cubing\,both\,sides,\\=>{{(p-2q)}^{3}}={{4}^{3}}\\=>{{p}^{3}}-3.{{p}^{2}}.2q+3.p.4{{q}^{2}}-8{{q}^{3}}=64\\=>{{p}^{3}}-6pq(p-2q)-8{{q}^{3}}=64\\=>{{p}^{3}}-24pq-8{{q}^{3}}-64=0\end{array}$

$ \displaystyle \begin{array}{l}{{\left( x+\frac{1}{2} \right)}^{2}}+{{q}^{2}}={{x}^{2}}+x+1\\=>{{x}^{2}}+\frac{1}{4}+x+{{q}^{2}}={{x}^{2}}+x+1\\=>{{q}^{2}}=\frac{3}{4}\\=>q=\pm \sqrt{\frac{3}{4}}=\pm \frac{1}{2}\sqrt{3}\end{array}$

$ \displaystyle \begin{array}{l}{{a}^{2}}-2a-1=0\\{{a}^{2}}-1=2a\\on\,dividing\,with\,a\\a-\frac{1}{a}=2\\Now\,\,{{a}^{2}}+\frac{1}{{{a}^{2}}}+3a-\frac{3}{a}\\{{\left( a-\frac{1}{a} \right)}^{2}}+2+3\left( a-\frac{1}{a} \right)\\4+2+3(2)\\4+2+6=12\end{array}$

\[\begin{align} & {{a}^{2}}+1=a \\ & =>{{a}^{2}}-a+1=0 \\ & Multiply\,both\,sides\,by\,(a+1), \\ & =>(a+1)({{a}^{2}}-a+1)=0 \\ & =>{{a}^{3}}+1=0 \\ & =>{{a}^{3}}=\,-1 \\ & Thus, \\ & {{a}^{12}}+{{a}^{6}}+1=1+1+1=3 \\ & \\ & Alternate\,solution \\ & {{a}^{2}}+1=a \\ & Cubing\,both\,sides \\ & =>{{a}^{6}}\,+\,1+\,3{{a}^{2}}\,({{a}^{2}}\,+\,1)\,=\,{{a}^{3}} \\ & =>{{a}^{6}}\,+\,1+\,3{{a}^{2}}\,(a)\,=\,{{a}^{3}}\,\,(initial\,condition\,used) \\ & =>{{a}^{6}}\,+\,1+\,3{{a}^{3}}\,=\,{{a}^{3}} \\ & =>{{a}^{6}}\,+\,1+\,2{{a}^{3}}\,\,=\,0 \\ & =>\,\,{{({{a}^{3}}\,+\,1)}^{2}}\,=\,0 \\ & =>\,\,{{a}^{3}}\,\,=\,-1 \\ \end{align}\] Using this, we can find the value of the expression. Remember, in this case, only one real root (a=-1) exists whereas the other two roots are imaginary in nature.