Question

NO SPAMS answer this pls​

Answer 1

Question 1 :-

Factorise 16a⁴ + 81b⁴ + 36a²b²

(4a²)² + (9b²)² + (6ab)²

By using identity, (a+b)² = a² + 2ab + b²

(4a²+9b²) = 16a⁴ + 81b⁴ + 72a²b²

Therefore, the question can be expressed as :

(4a²)² + (9b)² + 72a²b² - 36a²b²

(4a²+9b²)² - 36a²b²

Identity to use : a²-b² = (a+b)(a-b)

(4a²+9b²)² - (6ab)²

[(4a²+9b²)+6ab][(4a²+9b²)-6ab]

Question 2 :-

By using splitting the middle term method,

28m² + 25mn - 8n²

28m² - 7mn + 32mn + 8n²

7m(4m - n) + 8n(4m - n)

(4m - n)(7m + 8n)

Some more identities :-

(a+b)² = a² + 2ab + b²

(a-b)² = a² - 2ab + b²

a²-b² = (a-b)(a+b)

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

(x+a)(x+b) = x² + x(a+b) + ab

(a+b)³ = a³ + 3a²b + 3ab² + b³

a³+b³ = (a+b)(a²-ab+b²)

(a-b)³ = a³ - 3a²b + 3ab² - b³

a³-b³ = (a-b)(a²+ab+b²)

a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² - ab - bc - ca)

Conditional identity:

if a+b+c = 0,

a³+b³+c³ = 3abc