Question

factorise the following (I) 27y³ + 125z³ , (ii) 64m³ - 343m³

Answer 1

$\mathbb{ SOLUTION }$ :

(i) 27y³ + 125z³

We know that,
x³ + y³ = (x+y) ( x² - xy + y² )
So, here
x = 3y , y = 5z

= 27y³ + 125z³ = (3y + 5z) { (3y)² - 3y*5x + (5z)² }

= 27y³ + 125z³ = (3y + 5z) { 9y² - 15xy + 25z² }

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(ii) 64m³ - 343n³

We know that,
x³ - y³ = (x-y) ( x² - xy + y² )
So, here
x = 4m , y = 7n

= 64m³ - 343n³ = (4m - 7n) { (4m)² + 4m*7n + (7n)² }

= 64m³ - 343n³ = (4m - 7n) { 16m² + 28mn + 49n² }

Answer 2

$\blue{\huge{Solution}}$

I)
$27 {y}^{3} + 125 {z}^{3} \\ using \: identity \: {a}^{3} + {b}^{3} = (a + b)( {a}^{2} + {b}^{2} - ab) \\ a = 3y \\ b = 5z \\ 27 {y}^{3} + 125 {z}^{3} = (3y + 5z)(9 {y}^{2} + 25 {z}^{2} - 15yz)$

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II)
$64 {m}^{3} - 343 {n}^{3} \\ {(4m)}^{3} - {(7n)}^{3} \\ using \: identity - \\ {(a)}^{3} - {(b)}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab) \\ a = 4m \\ b = 7n \\ 64 {m}^{3} - 343 {n}^{3} = (4m - 7n)(16 {m}^{2} + 49 {n}^{2} + 28mn)$
Hope it will help you‼️‼️