Math, asked by ashfaquesoorya, 2 days ago

factorise
27p3-125q3​

Answers

Answered by LEGEND778
2

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)

Answered by hukam0685
0

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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