Answer :
To factorise the given expression
[tex]$ \frac{1}{27} p^3-8 q^3, $[/tex]
we can recognize that this is a difference of cubes, which can be generally expressed as \( a^3 - b^3 \). The formula for factoring a difference of cubes is:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
In our given expression:
[tex]\[ \frac{1}{27}p^3 - 8q^3, \][/tex]
we can rewrite it in a form similar to \( a^3 - b^3 \):
[tex]\[ \left(\frac{p}{3}\right)^3 - (2q)^3. \][/tex]
So, we identify:
[tex]\[ a = \frac{p}{3} \][/tex]
and
[tex]\[ b = 2q. \][/tex]
Using the difference of cubes formula \( (a^3 - b^3) = (a - b)(a^2 + ab + b^2) \), we substitute \( a = \frac{p}{3} \) and \( b = 2q \):
[tex]\[ \left(\frac{p}{3}\right)^3 - (2q)^3 = \left(\frac{p}{3} - 2q\right)\left(\left(\frac{p}{3}\right)^2 + \left(\frac{p}{3}\right)(2q) + (2q)^2\right). \][/tex]
Now let's simplify each part inside the parentheses:
1. The first term: \( a - b \)
[tex]\[ \frac{p}{3} - 2q. \][/tex]
2. The second term: \( a^2 + ab + b^2 \)
[tex]\[ \left(\frac{p}{3}\right)^2 + \left(\frac{p}{3}\right)(2q) + (2q)^2. \][/tex]
Simplify each part:
[tex]\[ \left(\frac{p}{3}\right)^2 = \frac{p^2}{9}, \][/tex]
[tex]\[ \left(\frac{p}{3}\right)(2q) = \frac{2pq}{3}, \][/tex]
[tex]\[ (2q)^2 = 4q^2. \][/tex]
Combining these together, we have:
[tex]\[ \frac{p^2}{9} + \frac{2pq}{3} + 4q^2. \][/tex]
Therefore, putting it all together, the factorised form of the given expression is:
[tex]\[ 8\left(\frac{p}{6} - q\right)\left(\frac{p^2}{36} + \frac{pq}{6} + q^2\right). \][/tex]
Simplifying the constant factor \( 8 \) out of the brackets:
[tex]\[ 8 \cdot \left( \frac{1}{6} \left(p - 6q\right) \right)\left(\frac{1}{36}\left(p^2 + 6pq + 36q^2\right)\right). \][/tex]
[tex]\[ 8 \cdot \frac{1}{6} \left(p - 6q\right) \cdot \frac{1}{36} \left(p^2 + 6pq + 36q^2\right). \][/tex]
Finally simplify:
[tex]\[ 8 \cdot \left(\frac{1}{6} (p - 6q)\right)\left(\frac{1}{36} (p^2 + 6pq + 36q^2)\right). \][/tex]
In its simplest form, we have:
[tex]\[ 8 \left(\frac{1}{6}p - q\right) \left(\frac{1}{36}p^2 + \frac{1/6}pq + q^2\right), \][/tex]
or more succinctly:
[tex]\[ 8 \left(\frac{1}{6}p - q\right) \left(0.027778p^2 + 0.166667pq + q^2\right). \][/tex]
[tex]$ \frac{1}{27} p^3-8 q^3, $[/tex]
we can recognize that this is a difference of cubes, which can be generally expressed as \( a^3 - b^3 \). The formula for factoring a difference of cubes is:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
In our given expression:
[tex]\[ \frac{1}{27}p^3 - 8q^3, \][/tex]
we can rewrite it in a form similar to \( a^3 - b^3 \):
[tex]\[ \left(\frac{p}{3}\right)^3 - (2q)^3. \][/tex]
So, we identify:
[tex]\[ a = \frac{p}{3} \][/tex]
and
[tex]\[ b = 2q. \][/tex]
Using the difference of cubes formula \( (a^3 - b^3) = (a - b)(a^2 + ab + b^2) \), we substitute \( a = \frac{p}{3} \) and \( b = 2q \):
[tex]\[ \left(\frac{p}{3}\right)^3 - (2q)^3 = \left(\frac{p}{3} - 2q\right)\left(\left(\frac{p}{3}\right)^2 + \left(\frac{p}{3}\right)(2q) + (2q)^2\right). \][/tex]
Now let's simplify each part inside the parentheses:
1. The first term: \( a - b \)
[tex]\[ \frac{p}{3} - 2q. \][/tex]
2. The second term: \( a^2 + ab + b^2 \)
[tex]\[ \left(\frac{p}{3}\right)^2 + \left(\frac{p}{3}\right)(2q) + (2q)^2. \][/tex]
Simplify each part:
[tex]\[ \left(\frac{p}{3}\right)^2 = \frac{p^2}{9}, \][/tex]
[tex]\[ \left(\frac{p}{3}\right)(2q) = \frac{2pq}{3}, \][/tex]
[tex]\[ (2q)^2 = 4q^2. \][/tex]
Combining these together, we have:
[tex]\[ \frac{p^2}{9} + \frac{2pq}{3} + 4q^2. \][/tex]
Therefore, putting it all together, the factorised form of the given expression is:
[tex]\[ 8\left(\frac{p}{6} - q\right)\left(\frac{p^2}{36} + \frac{pq}{6} + q^2\right). \][/tex]
Simplifying the constant factor \( 8 \) out of the brackets:
[tex]\[ 8 \cdot \left( \frac{1}{6} \left(p - 6q\right) \right)\left(\frac{1}{36}\left(p^2 + 6pq + 36q^2\right)\right). \][/tex]
[tex]\[ 8 \cdot \frac{1}{6} \left(p - 6q\right) \cdot \frac{1}{36} \left(p^2 + 6pq + 36q^2\right). \][/tex]
Finally simplify:
[tex]\[ 8 \cdot \left(\frac{1}{6} (p - 6q)\right)\left(\frac{1}{36} (p^2 + 6pq + 36q^2)\right). \][/tex]
In its simplest form, we have:
[tex]\[ 8 \left(\frac{1}{6}p - q\right) \left(\frac{1}{36}p^2 + \frac{1/6}pq + q^2\right), \][/tex]
or more succinctly:
[tex]\[ 8 \left(\frac{1}{6}p - q\right) \left(0.027778p^2 + 0.166667pq + q^2\right). \][/tex]
