if x-y=3, xy= 10 find the value of x^3-y^3
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Hey!!!....Here is ur answer
x^3-y^3=(x-y)(x^2+xy+y^2)
x-y=3
squaring both sides
x^2+y^2-2xy=9
x^2+y^2-2×10=9
x^2+y^2=29
Now x^3-y^3=(x-y)(x^2+xy+y^2)
=3 (29+10)
=3×39
=117
So x^3-y^3=117
Hope it will help you
x^3-y^3=(x-y)(x^2+xy+y^2)
x-y=3
squaring both sides
x^2+y^2-2xy=9
x^2+y^2-2×10=9
x^2+y^2=29
Now x^3-y^3=(x-y)(x^2+xy+y^2)
=3 (29+10)
=3×39
=117
So x^3-y^3=117
Hope it will help you
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