Answer:
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Step-by-step explanation:
According to the identity if a+b+c=0
then a3+b3+c3=3abc
a3+b3+c3/abc=3
a2*a/bc*a+b2*b/ca*b+c2*c/ab*c=3
cancel a,b,c in all the fraction then you get
<span>a²/bc+b²/ca+c²/ab=3.
</span>hence proved
Let the fist integer be x, the second is x+20
the product of the numbers is:
x(x+20)
the sum of the numbers is:
x+x+20=2x+20
the sum of the above operations will give us:
2x+20+x^2+20x=95
x^2+22x+20=95
this can be written as quadratic to be:
x^2+22x-75=0
solving the above we get:
x=3 and x=-25
but since the integers should be positive, then x=3
the second number is x+20=3+20=23
hence the numbers are:
3 and 23
