A) 5
B) 6
C) 7
D) 8
Correct Answer: C
Solution :
We know that \[{{(a+b)}^{5}}+{{(a-b)}^{5}}\] \[={{\,}^{5}}{{C}_{0}}{{a}^{5}}+{{\,}^{5}}{{C}_{1}}{{a}^{4}}b+{{\,}^{5}}{{C}_{2}}{{a}^{3}}{{b}^{2}}\] \[+{{\,}^{5}}{{C}_{3}}{{a}^{2}}{{b}^{3}}+{{\,}^{5}}{{C}_{4}}a{{b}^{4}}+{{\,}^{5}}{{C}_{5}}{{b}^{5}}\] \[+{{\,}^{5}}{{C}_{0}}{{a}^{5}}-{{\,}^{5}}{{C}_{1}}{{a}^{4}}b+{{\,}^{5}}{{C}_{2}}{{a}^{3}}{{b}^{2}}\] \[-{{\,}^{5}}{{C}_{3}}{{a}^{2}}{{b}^{3}}{{+}^{5}}{{C}_{4}}a{{b}^{4}}-{{\,}^{5}}{{C}_{5}}{{b}^{5}}\] \[=2[{{a}^{5}}+10{{a}^{3}}{{b}^{2}}+10a{{b}^{4}}]\] \[\therefore \] \[{{[x+{{({{x}^{3}}-1)}^{1/2}}]}^{5}}+{{[x-{{({{x}^{3}}-1)}^{1/2}}]}^{5}}\] \[=2[{{x}^{5}}+10{{x}^{3}}({{x}^{3}}-1)+10x{{({{x}^{3}}-1)}^{2}}]\] Therefore, the given expression is a polynomial of degree 7.
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