Question

factorise;
27a^3+8b^3+54a^2b+36ab^2+729

Answer 1

Solution :-

→ 27a^3 + 8b^3 + 54a^2b + 36ab^2 + 729

→ (27a^3 + 8b^3 + 54a^2b + 36ab^2) + 729

→ [(3a)³ + (2b)³ + 3*9a²*2b + 3*3a*4b²] + 729

→ [(3a)³ + (2b)³ + 3*(3a)²*2b + 3*3a*(2b)²] + 729

using x³ + y³ + 3x²y + 3xy² = (x + y)³ .

so,

• x = 3a
• y = 2b .

then,

→ [(3a)³ + (2b)³ + 3*(3a)²*2b + 3*3a*(2b)²] + 729

→ (3a + 2b)³ + (9)³

now, using x³ + y³ = (x + y)(x² + y² - xy)

→ (3a + 2b + 9)[(3a + 2b)² + 9² - 9(3a + 2b)]

→ (3a + 2b + 9)[9a² + 4b² + 12ab + 81 - 27a - 18b]

→ (3a + 2b + 9)[9a² + 4b² + 12ab - 27a - 18b + 81]

Learn more :-

JEE mains Question :-

https://brainly.in/question/22246812

. Find all the zeroes of the polynomial x4

– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.

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