Question

please help me...!!!
solve question no. 11​

Answer 1

Answer:

YOUR QUESTION

Factorise: 27x3 + y + z3 – 9xyz

= (3x)3 + (y)3 + (z)3 – 9xyz

= (3x)3 + (y)3 + (z) – 3 × 3x × y × z

= Using identity :

{a3 + b3 + c3 – 3abc = (a + b + c)(a + b2 + c2 - ab – bc - ca) putting a = 3x , b=y , c = z

= (3x + y +z) (9x² + y2 + z² - 3xy - yz - 3zx)

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Answer 2

Question :

Factorise 27x³ + y³ + z³ - 9xyz

Answer :

27x³ + y³ + z³ -9xyz

This can be written in the form :

• (3x)³ + y³ + z³ - 3 × (3x) × y × z

Now , if we compare it to identity

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

• a = 3x
• b= y
• c = z

•°• , 27x³ + y³ + z³ - 9xyz

= ( 3x + y +z)[(3x)² + y² + z² - 3xy - yz - z3x]

= (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx)

Special Case :

• If a + b + c = 0 ,

• •°• , a³ + b³ + c³ = 3abc