Question

Factorize each of the following:
(i) \[8{a^3} + {b^3} + 12{a^2b} + 6a{b^2}\]
(ii) \[8{a^3} - {b^3} - 12{a^2}b + 6a{b^2}\]
(iii) \[27 - 125{a^3} - 135a + 225{a^2}\]
(iv) \[64{a^3} - 27{b^3} - 144{a^2}b + 108a{b^2}\]
 (v) $27{p^3} - \dfrac{1}{{216}} - \dfrac{9}{2}{p^2} + \dfrac{1}{4}p$

Answer 1

Hint:To solve this question use the Basic algebraic identities and formulas i.e ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$ and ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)$.Compare these standard formulas with given equations and get the answer.

Complete step-by-step answer:
Take the first equation:
1. \[8{a^3} + {b^3} + 12{a^2b} + 6a{b^2}\]
this can be written as:
$ = {\left( {2a} \right)^3} + {b^3} + 6ab\left( {2a + b} \right)$....taking as equation $(1)$
Comparing equation $(1)$ with ${a^3} + {b^3} + 3ab\left( {a + b} \right) = {\left( {a + b} \right)^3}$
therefore equation will be:
$
   = {\left( {2a + b} \right)^3} \\
   = \left( {2a + b} \right)\left( {2a + b} \right)\left( {2a + b} \right) \\
$
Taking the second equation:
2. \[8{a^3} - {b^3} - 12{a^2}b + 6a{b^2}\]
this can be written as:
$ = {\left( {2a} \right)^3} - {b^3} - 6ab\left( {2a - b} \right)$....taking as equation $(2)$
Comparing equation $(2)$ with ${a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( {a - b} \right)^3}$
therefore equation will be:
$
   = {\left( {2a - b} \right)^3} \\
   = \left( {2a - b} \right)\left( {2a - b} \right)\left( {2a - b} \right) \\
$
Taking equation third:
3. \[27 - 125{a^3} - 135a + 225{a^2}\]
this can be written as:
\[ = {\left( 3 \right)^3} - {\left( {5a} \right)^3} - 45a\left( {3 - 5a} \right)\]....taking as equation $(3)$
Comparing equation $(3)$ with ${a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( {a - b} \right)^3}$ equation will be:
$
   = {\left( {3 - 5a} \right)^3} \\
   = \left( {3 - 5a} \right)\left( {3 - 5a} \right)\left( {3 - 5a} \right) \\
$
Taking equation fourth
4. \[64{a^3} - 27{b^3} - 144{a^2}b + 108a{b^2}\]
this can be written as
$ = {\left( {4a} \right)^3} - {\left( {3b} \right)^3} - 36ab\left( {4a - 3b} \right)$...taking as equation $(4)$
Comparing equation $(4)$ with ${a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( {a - b} \right)^3}$ equation will be:
$
   = {\left( {4a - 3b} \right)^3} \\
   = \left( {4a - 3b} \right)\left( {4a - 3b} \right)\left( {4a - 3b} \right) \\
$
Taking equation fifth:
4. $27{p^3} - \dfrac{1}{{216}} - \dfrac{9}{2}{p^2} + \dfrac{1}{4}p$
this can be written as:
$ = \left( {3{p^3}} \right) - {\left( {\dfrac{1}{6}} \right)^3} - 3 \times 3p \times \dfrac{1}{6}\left( {3p - \dfrac{1}{6}} \right)$...taking as equation $(5)$
Comparing equation $(5)$ ${a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( {a - b} \right)^3}$ equation will be:
$
   = {\left( {3p - \dfrac{1}{6}} \right)^3} \\
   = \left( {3p - \dfrac{1}{6}} \right)\left( {3p - \dfrac{1}{6}} \right)\left( {3p - \dfrac{1}{6}} \right) \\
$
Hence we got the desired results.

Note:- In this question all the given five polynomial equations are in the cubic form hence we factorized them using algebraic cubic formula and after simplification got the desired results for each of them.Students should remember Basic algebraic identities of quadratic and cubic forms for solving these types of questions.