Math, asked by ursrandhir2242, 1 year ago

if x^2+y^2=90 and xy=27 ,then find the value of x^3-y^3 where x>y.

Answers

Answered by rakeshjizbgmailcom

0

x^3-y^3=(x-y)(x^2+y^2+xy)
=(6)(90+27)
=6*117
=702

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