Question

27a3+64b3 factorise​

Answer 1

$27 {a}^{3} + 64 {b}^{3} \\ = (3a + 4b)(9 {a}^{2} - 12ab + 16 {b}^{2} )$

Answer 2

Answer:

(3a + 4b)(9a² - 12ab + 16b²)

Step-by-step explanation:

27a³+64b³

⇒ (3a)³ + (4b)³

As we know,

• x³ + y³ = (x + y)(x² - xy + y²)

• Here, 3a = x and 4b = y

∴(3a)³ + (4b)³ = [(3a) + (4b)][(3a)² - (3a)(4b) + (4b)²]

⇒(3a + 4b)(9a² - 12ab + 16b²)

Answer⇒ (3a + 4b)(9a² - 12ab + 16b²)

More things to know:-

• (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 + 2 ca

                          = a² + b² + c² + 2(ab + bc + ca)  

• (a + b)³ = a³ + b³ + 3ab(a + b)

                    =  a³ + b³ + 3a²b + 3ab²  

• (a - b)³ = a³ - b³ - 3ab(a - b)                    

                   = a³ - b³ - 3a²b + 3ab²    

• a³ + b³ = (a + b)(a² - ab + b²)

• a³ - b³ = (a - b)(a² + ab + b²)

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

• (x + p)(x + q) = x² + (p + q)x + pq

I hope that this helped

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