Question

The expansion of ${(x+\sqrt{x^3-1})}^5+{(x-\sqrt{x^3-1})}^5$ is a polynomial of degree

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

$=(x+\sqrt{x^3-1}{)}^5+(x-\sqrt{x^3-1}{)}^5$

$\Rightarrow (a+b{)}^5+$ $(ab{)}^5=a^5{+}^5{c}_1{a}^{4}b{+}^5{c}_2{a}^3{b}^2{+}^5{c}_3{a}^2{b}^3+$ ${}^5{c}_{4}a{b}^4+b^5+a^5$ ${}^{-5}{c}_1{a}^{4}b{+}^5{c}_2{a}^3{b}^2-$ ${}^5{c}_3{a}^2{b}^3{+}^5{c}_{4}a{b}^4-b^5$

$=2a^5+{2.}^5{c}_2{a}^3{b}^2+{2.}^5{c}_{4}a{b}^4$

$a=x$ $b=\sqrt{x^3-1}$

$(x+\sqrt{x^3-1}{)}^5+(x-\sqrt{x^3-1}{)}^5=2x^5+{2.}^5{c}_2{x}^3(x^3-1)+{2.}^5{c}_4\left(x\right)(x^3-1{)}^2$

$=2x^5+{2.}^5{c}_2(x^6-x^3)+{2.}^5{c}_4(x^7+x-2x^4)$

So :- this polynomial has degree $7$.