Question

Factorise :

$p^3-10+\frac{27}{p^3}$

• A. $(p+2+\frac{3}{p})(p^2+4+\frac{9}{p^2}-2p-\frac{6}{p}-3)$
• B. $(2p+2+\frac{3}{p})(p^2+4+\frac{9}{p^3}-2p-\frac{6}{p}-3)$
• C. $(2p+1+\frac{3}{p})(p^2+4+\frac{9}{p^2}-2p-\frac{7}{p}-3)$
  
• D. $(p-2+\frac{3}{p})(p^2+4+\frac{9}{p^2}-3p-\frac{7}{p}-3)$

Answer 1

The correct option is A $(p+2+\frac{3}{p})(p^2+4+\frac{9}{p^2}-2p-\frac{6}{p}-3)$

Given, $p^3-10+\frac{27}{p^3}$
Using, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Here,
$a=p$
$b=2$
$c=\frac{3}{p}$,
Therefore,
$p^3+2^3+(\frac{3}{p}{)}^3-3\times p\times 2\times \frac{3}{p}$
$=(p+2+\frac{3}{p})(p^2+4+\frac{9}{p^2}-2p-\frac{6}{p}-3)$