factorize 27x^3+y^3+z^3-9xyz using identity
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Hey friend...
Here is your answer
27x3 + y3 + z3 - 9xyz
= (3x)3 + y3 + z3 - 3 × (3x) × y × z
= [3x + y + z ][(3x)2 + y2 + z2 - (3x) × y − y × z − z(3x)] [Since (a + b + c )(a2 + y2 + z2 - ab−bc−ca)]
= [3x + y + z ][9x2 + y2 + z2 - 3xy − yz − 3zx
I hope it will help you
#yahyaahmad#
Here is your answer
27x3 + y3 + z3 - 9xyz
= (3x)3 + y3 + z3 - 3 × (3x) × y × z
= [3x + y + z ][(3x)2 + y2 + z2 - (3x) × y − y × z − z(3x)] [Since (a + b + c )(a2 + y2 + z2 - ab−bc−ca)]
= [3x + y + z ][9x2 + y2 + z2 - 3xy − yz − 3zx
I hope it will help you
#yahyaahmad#
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