Question

factorise using algebraic identities (1to9)
factorisation ​

Answer 1

Answer:

i have answered a few questions only

Answer 2

Solution :

4 (I) -

16a² - 25/9a²

=> [ 4a ]² - [ 5/3a ]²

=> [ 4a + 5/3a ][ 4a - 5/3a ]

=> [ 12a² + 5 ][ 12a² - 5] / 9a² .

4 (II) -

⅑ x²y² - 1/25 y²z²

=> [⅓ xy]² - [ ⅕ yz ]²

=> [ ⅓ xy + ⅕ yz ][ ⅓ xy - ⅕ yz ]

=> y² [ ⅓ x + ⅕ y][ ⅓ x - ⅕ y ]

4(III) -

x⁸ - 1

=> [ x⁴ ]² - 1

=> [ x⁴ + 1 ][ x² - 1]

=> [ x⁴ + 1][ x + 1][ x - 1]

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Additional Information -

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

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