Question

FACTORISE : 125p^3-216q^3

Answer 1

Answer:

(5p - 6q) • (25p^2 + 30pq + 36q^2)

Step-by-step explanation:

STEP1:Equation at the end of step 1

(125 • (p^3)) - (2^3•3^3 q^3)

STEP2:Equation at the end of step2

5^3 p^3 - (2^3•3^3 q3)

STEP3:

Trying to factor as a Difference of Cubes

3.1 Factoring: 125p^3-216q^3

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

(a-b) • (a^2 +ab +b^2)

Proof : (a-b)•(a^2+ab+b^2) =

a^3+a^2b+ab^2-ba^2-b^2a-b^3 =

a^3+(a^2b-ba^2)+(ab^2-b^2a)-b^3 =

a^3+0+0+b^3 =

a^3+b^3

Check : 125 is the cube of 5

Check : 216 is the cube of 6

Check : p3 is the cube of p1

Check : q3 is the cube of q1

Factorization is :

(5p - 6q) • (25p^2 + 30pq + 36q^2)

Trying to factor a multi variable polynomial :

3.2 Factoring 25p^2 + 30pq + 36q^2

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Final result :

(5p - 6q) • (25p^2 + 30pq + 36q^2)

I hope it's helpful for you..☺️

Answer 2

Step-by-step explanation:

Given: $125 {p}^{3} -216 {q}^{3}$

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.

$125 {p}^{3} -216 {q}^{3} = ( {5p)}^{3} - {(6q)}^{3}$

Step 2: Apply identity to factorise.

Here, we can write that $x = 5p$ and $y = 6q$

$( {5p)}^{3} - {(6q)}^{3} = (5p -6 q)( {(5p)}^{2} + 5p(6q) + {(6q)}^{2} )$

simply

$( {5p)}^{3} - {(6q)}^{3} = (5p -6q)( {25p}^{2} + 30pq + {36q}^{2} )$

Final answer:

$\bf {125p}^{3} - {216q}^{3} = (5p -6q)( {25p}^{2} + 30pq + {36q}^{2} )$

Hope it helps you.

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