Question

One of the factors of $p^6+4p^3–1$ is

• A. $p^4+p^3+2p^2+p+1$
• B. $p^2+p+1$
• C. $p^2+p–1$
• D. $p^4–p^3–2p^2+p+1$

Answer 1

The correct option is C $p^2+p–1$

$p^6+4p^3–1=p^6+3p^3+p^3–1$
$=\left[\right(p^2{)}^3+p^3+(-1{)}^3]+3p^3$
$=\left[\right(p^2{)}^3+p^3+(-1{)}^3]-3\times p^2\times p\times (-1)$
Using the identity , $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$, we get
$(p^2{)}^3+p^3+(-1{)}^3-3\times p^2\times p\times (-1)$
$=(p^2+p-1)\left(\right(p^2{)}^2+p^2+(-1{)}^2-p^3+p+p^2)$
$=(p^2+p-1)(p^4+p^2+1-p^3+p+p^2)$
$=(p^2+p-1)(p^4-p^3+2p^2+p+1)$
Thus, $(p^2+p–1)$ is one of the factor of $p^6+4p^3–1$.
Hence, the correct answer is option (3).