factorise 8xcube + ycube + 2cube _6xyz using identity
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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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