Math, asked by simhaa73, 1 month ago

factorise 8xcube + ycube + 2cube _6xyz using identity​

Answers

Answered by Suhanmasterblaster
1

Step-by-step explanation:

8x³ +y³ + z³ - 6xyz

= (2x)³ +y³ + z³ - 3(2x)(y)(z)

Now using identity : a^{3}\ +\ b^{3}\ +\ c^{3}\ -\ 3abc\ =\ (a+b+c)(a^{2}\ +\ b^{2}\ +\ c^{2}\ -\ ab-bc-ca)a

3

+ b

3

+ c

3

− 3abc = (a+b+c)(a

2

+ b

2

+ c

2

− ab−bc−ca)

⇒ (2x+y+z)[(2x)² + y² + c² - 2xy - yz - 2xz]

= (2x+y+z)(4x² + y² + c² - 2xy - yz - 2xz)

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