Question

What is the square root of $\left(\frac{x^5-1}{x-1}\right)+(x^3+2x^2+x)$?

• A. $x^2+x+1$
• B. $x^2-x+1$
• C. $x^2-x-1$
• D. $x^2+x-1$

Answer 1

The correct option is A $x^2+x+1$

$\frac{x^5-1}{x-1}+x^3+2x^2+x$
$=\frac{x^5-1+x^4+2x^3+x^2-x^3-2x^2-x}{x-1}$
$=\frac{x^5+x^4+x^3-x^3-x-1}{x-1}=\frac{x^3(x^3+x+1)-1(x^2+x+1)}{(x-1)}$
$=\frac{(x^2+x+1)(x^3-1)}{x-1}=\frac{(x^2+x+1)(x-1)(x^2+x+1)}{(x-1)}$
$=(x^2+x+1{)}^2$
$\therefore$ Required square root $=\sqrt{(x^2+x+1{)}^2}=(x^2+x+1)$.