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)

7

Answer 2

We know that,
$(a+b)^5+(a-b)^5= \, ^5C_0 a^5+ \, ^5C_1 a^4b+ \, ^5C_2 a^3b^2$
$\, \, \, \, \, + ^5C_3a^2b^3+ \, ^5C_4ab^4+ \, ^5C_5b^5+^5C_0 a^5- \, ^5C_1 a^4b$
$\, \, \, \, \, \, \, \, \, \, \, \, \, + ^5 C_2 a^3b^2- \, ^5C_4 ab^4+ \, ^5C_3 a^2b^3+ \, ^5C_4 ab^4- \, ^5C_5 b^5$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =2[a^5+10a^3+b^2+5ab64]$
$\therefore \, [x+(x^3-1)^{1/5}]^5 + [ x-(x^3-1)^{1/2}]^5$
=2 $[x^5+10x^3(x^3-1)+5x(x^3-1)^2]$
Therefore, the given expression is a polynomial of degree 7.