Question

"Question 10 Factorise each of the following:
(i) 27y^3 + 125z^3
(ii) 64m^3 - 343n^3
[Hint: See question 9.]

Class 9 - Math - Polynomials Page 49"

Answer 1

Solution:

(i) 27y³ + 125z³

Using identity,

x³ + y³ = (x + y) (x² – xy + y²)

27y³ + 125z³

= (3y)³ + (5z)³

= (3y + 5z) {(3y)² – (3y)(5z) + (5z)²}

= (3y + 5z) (9y² – 15yz + 25z²)

 

(ii) 64m³ – 343n³

Using identity,

x³ – y³= (x – y) (x² + xy + y² )

64m³– 343n³

=(4m)³ – (7n)³

=(4m - 7n) {(4m)² + (4m)(7n) + (7n)²}

= (4m-7n) (16m²+ 28mn + 49n²)

 

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Hope this will help you.....

Answer 2

Hi

Here is your answer:)

I) Given us that

27y³ + 125z³

We can also write this way,

( 3y )³ + ( 5z )³

we know that ,

a³ + b³ = ( a + b ) ( a² - ab + b² )

so,

( 3y )³ + ( 5z )³

= ( 3y + 5z ) { ( 3y )² - ( 3y ) . ( 5z ) + ( 5z )² }

= ( 3y + 5z ) ( 9y² - 15yz + 25z² )

=================================

ii )

Given us that

64m³ - 343n³

We can also write this way,

( 4m )³ - ( 7n )³

we know that, a³ - b³ = ( a - b ) ( a + ab + b )
.so,

((4m)³ - ( 7n )³ = ( 4m - 7n ) ((4m )²+ ( 4m ) ( 7n ) + ( 7n )²)

= ( 4m - 7n ) ( 16m² + 28mn + 49 )

Hopes I helped

Answered by a Useless Benefactor