A) \[{{({{a}^{2}}+{{b}^{2}})}^{3}}\]
B) \[{{({{a}^{3}}+{{b}^{3}})}^{2}}\]
C) \[{{({{a}^{4}}+{{b}^{4}})}^{2}}\]
D) \[{{({{a}^{2}}+{{b}^{2}})}^{4}}\]
Correct Answer: B
Solution :
Let \[\Delta =\left| \begin{matrix} {{a}^{2}} & 2ab & {{b}^{2}} \\ {{b}^{2}} & {{a}^{2}} & 2ab \\ 2ab & {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right|\] Applying, \[{{C}_{1}}\to {{C}_{2}}+{{C}_{3}},\] we get \[\Rightarrow \] \[\Delta \left| \begin{matrix} {{(a+b)}^{2}} & 2ab & {{b}^{2}} \\ {{(a+b)}^{2}} & {{a}^{2}} & 2ab \\ {{(a+b)}^{2}} & {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right|\] \[={{(a+b)}^{2}}\left| \begin{matrix} 1 & 2ab & {{b}^{2}} \\ 1 & {{a}^{2}} & 2ab \\ 1 & {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right|\] Apply \[{{R}_{2}}\to {{R}_{1}}-{{R}_{1}},\,{{R}_{3}}\Rightarrow {{R}_{3}}-{{R}_{1}},\] we get \[\Delta ={{(a+b)}^{2}}\left| \begin{matrix} 1 & 2ab & {{b}^{2}} \\ 0 & {{a}^{2}}-2ab & 2ab-{{b}^{2}} \\ 0 & {{b}^{2}}-2ab & {{a}^{2}}-{{b}^{2}} \\ \end{matrix} \right|\] \[={{(a+b)}^{2}}\left| \begin{matrix} a(a-2b) & b(2a-b) \\ b(b-2a) & (a-b)\,(a+b) \\ \end{matrix} \right|\] \[={{(a+b)}^{2}}\{a(a-b)(a+b)(a-2b)\] \[-{{b}^{2}}(b-2a)\,(2a-b)\}\] \[={{(a+b)}^{2}}\{a({{a}^{2}}-{{b}^{2}})(a-2b)+{{b}^{2}}{{(2a-b)}^{2}}\}\] \[={{(a+b)}^{2}}\{({{a}^{2}}-{{b}^{2}})\,({{a}^{2}}-2ab)\] \[+{{b}^{2}}(4{{a}^{2}}+{{b}^{2}}-4ab)\}\] \[={{(a+b)}^{2}}\{{{a}^{4}}-{{a}^{2}}{{b}^{2}}-2{{a}^{3}}b+2a{{b}^{3}}\] \[+4{{a}^{2}}{{b}^{2}}+{{b}^{4}}-4a{{b}^{3}}\}\] \[={{(a+b)}^{2}}\{{{a}^{4}}+{{b}^{4}}+3{{a}^{2}}{{b}^{2}}-2{{a}^{3}}b-2a{{b}^{3}}\}\] \[={{(a+b)}^{2}}\{{{({{a}^{2}}+{{b}^{2}})}^{2}}-2ab({{a}^{2}}+{{b}^{2}})+{{a}^{2}}{{b}^{2}}\}\] \[={{(a+b)}^{2}}\{({{a}^{2}}+{{b}^{2}})({{a}^{2}}+{{b}^{2}}-2ab)+{{a}^{2}}{{b}^{2}}\}\] \[={{(a+b)}^{2}}\{({{a}^{2}}+{{b}^{2}}){{(a-b)}^{2}}+{{a}^{2}}{{b}^{2}}\}\] \[=({{a}^{2}}+{{b}^{2}})({{a}^{2}}-{{b}^{2}})+{{a}^{2}}{{b}^{2}}{{(a+b)}^{2}}\] \[=({{a}^{4}}-{{b}^{4}})({{a}^{2}}-{{b}^{2}})+{{a}^{2}}{{b}^{2}}{{(a+b)}^{2}}\] \[={{a}^{6}}-{{a}^{2}}{{b}^{4}}-{{a}^{4}}{{b}^{2}}+{{b}^{6}}+{{a}^{4}}{{b}^{2}}+{{a}^{2}}{{b}^{4}}+2{{a}^{3}}{{b}^{3}}\] \[={{a}^{6}}+2{{a}^{3}}{{b}^{3}}+{{b}^{6}}={{({{a}^{3}}+{{b}^{3}})}^{2}}\]
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