Math, asked by K2omangas7aurramz, 1 year ago

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


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Answers

Answered by abhi178
40
we know x^2+y^2-2xy=(x-y)^2
so, (x-y)^2=90-54=36
x-y=+6 (x> y)
now x^3-y^3=(x-y)(x^2+y^2+xy)
=6*(90+27)
=702
Answered by snas217236
4

Answer:

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