Question

The expression $[x+(x^3-1{)}^{1/2}{]}^5+[x-(x^3-1{)}^{1/2}{]}^5$ is a polynomial of degree

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (C)

We know that,
$(a+b{)}^5+(a-b{)}^5={}^5{C}_0{a}^5+{}^5{C}_1{a}^{4}b+{}^5{C}_2{a}^3{b}^2$
${+}^5{C}_3{a}^2{b}^3+{}^5{C}_{4}a{b}^4+{}^5{C}_5{b}^5{+}^5{C}_0{a}^5-{}^5{C}_1{a}^{4}b$
${+}^5{C}_2{a}^3{b}^2-{}^5{C}_{4}a{b}^4+{}^5{C}_3{a}^2{b}^3+{}^5{C}_{4}a{b}^4-{}^5{C}_5{b}^5$
$=2[a^5+10a^3+b^2+5a{b}^4]$
$\therefore [x+(x^3-1{)}^{1/5}{]}^5+[x-(x^3-1{)}^{1/2}{]}^5$
=2 $[x^5+10x^3(x^3-1)+5x(x^3-1{)}^2]$
Therefore, the given expression is a polynomial of degree $7$.