Question

Factorise the following :

$(p-3q{)}^3+(3q-7r{)}^3+(7r-p{)}^3$

• A. $3(p-3q)(q-r)(r-p)$
• B. $8(p-3q)(3q-r)(r-p)$
• C. $3(p-3q)(3q-7r)(7r-p)$
• D. $8(p-3q)(q-7r)(r+p)$

Answer 1

Answer: (C)

We know the algebraic identity is,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
If, $a+b+c=0$ then,
$\Rightarrow a^3+b^3+c^3-3abc=0(a^2+b^2+c^2-ab-bc-ca)$

$\Rightarrow a^3+b^3+c^3=0$

$\Rightarrow a^3+b^3+c^3=3abc$
Now,
$(p-3q{)}^3+(3q-7r{)}^3+(7r-p{)}^3$
Here,
$a=p-3q$
$b=3q-7r$
$c=7r-p$
Now,
$p-3q+3q-7r+7r-p=0$
Thus,
$(p-3q{)}^3+(3q-7r{)}^3+(7r-p{)}^3=3(p-3q\left)\right(3q-7r\left)\right(7r-p)$

Hence, Option C is correct.