factorise 27 a cube + b cube + c cube - 18abc using identity
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Hi ,
Plz , there is an error in the question.
****************************
We know the algebraic identity,
x³ + y³ + z³ - 3xyz
= ( x + y + z )( x² + y² + z² - xy - yz - zx )
********************************
Now ,
27a³ + b³ + c³ - 9abc
= ( 3a )³ + b³ + c³ - 3 × 3a × b × c
= (3a + b + c )[(3a)²+b²+c² -3ab-bc-3ac]
= (3a+b+c)[9a²+b²+c²-3ab-bc-3ac]
I hope this helps you.
: )
Plz , there is an error in the question.
****************************
We know the algebraic identity,
x³ + y³ + z³ - 3xyz
= ( x + y + z )( x² + y² + z² - xy - yz - zx )
********************************
Now ,
27a³ + b³ + c³ - 9abc
= ( 3a )³ + b³ + c³ - 3 × 3a × b × c
= (3a + b + c )[(3a)²+b²+c² -3ab-bc-3ac]
= (3a+b+c)[9a²+b²+c²-3ab-bc-3ac]
I hope this helps you.
: )
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