Question

factorize 2root2a³+ 3root3b³+ c³- 3 root 6ab​

Answer 1

Step-by-step explanation:

We are given that: (3√3a)³ + 27b³ - c³ + 9√3abc

we use the following identity,

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

we have,

(3√3a)³ + 27b³ - c³ + 9√3abc

= (3√3a)³ + (3b)³ + (-c)³ - 3 × 3√3a × 3b × (-c)

= ( 3√3a + 3b + (-c) ) ( (3√3a)² + (3b)² + (-c)² - (3√3a)(3b) - (3b)(-c) - (-c)(3√3a) )

= ( 3√3a + 3b - c ) ( 27a² + 9b² + c² - 9√3ab + 3cb + 3√3ac )

Therefore, (3√3a)³ + 27b³ - c³ + 9√3abc = ( 3√3a+3b-c )( 27a²+9b²+c²-9√3ab+3cb+3√3ac )