Question

factorise
27p3-125q3​

Answer 1

Answer:

(3p - 5q)  •  (9p2 + 15pq + 25q2)

Step-by-step explanation:

Factoring:  27p3-125q3  

Theory : A difference of two perfect cubes,  a3 - b3 can be factored into

             (a-b) • (a2 +ab +b2)

Proof :  (a-b)•(a2+ab+b2) =

           a3+a2b+ab2-ba2-b2a-b3 =

           a3+(a2b-ba2)+(ab2-b2a)-b3 =

           a3+0+0+b3 =

           a3+b3

Check :  27  is the cube of  3  

Check :  125  is the cube of   5  

Check :  p3 is the cube of   p1

Check :  q3 is the cube of   q1

Factorization is :

            (3p - 5q)  •  (9p2 + 15pq + 25q2)

Answer 2

Step-by-step explanation:

Given: $27 {p}^{3} -125 {q}^{3}$

To find: Factorise.

Solution:

Identity used:

$\boxed{\bf {x}^{3} - {y}^{3} = (x -y)( {x}^{2} +xy + {y}^{2} ) }$

Step 1: Convert given polynomial in standard form.

$27 {p}^{3} -125 {q}^{3} = ( {3p)}^{3} - {(5q)}^{3}$

Step 2: Apply identity to factorise.

$( {3p)}^{3} - {(5q)}^{3} = (3p -5 q)( {(3p)}^{2} + 3p(5q) + {(5q)}^{2} )$

simply

$( {3p)}^{3} - {(5q)}^{3} = (3p -5q)( {9p}^{2} + 15pq + {25q}^{2} )$

Final answer:

$\bf {27p}^{3} - {125q}^{3} = (3p -5q)( {9p}^{2} + 15pq + {25q}^{2} )$

Hope it helps you.

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