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x + y = 7 &
x^3 + y^3 = 133
Where x,y both are real numbers.
x is 5
y is 2
x^3 + y^3 = 133
(x+y)(x^2 + y^2 - xy) = 133
(7)(x^2 + y^2 - xy + 2xy - 2xy) = 133
(7)( (7*7) - 3xy ) = 133
343 - 21xy = 133
343 - 133 = 21xy
210 = 21xy
xy = 10
Ans :- 10
another way of solving is x^3+y^3=(x+y)(x^2-xy+y^2) substitute
133=7*(x^2+xy+y^2) or (x^2-xy+y^2)=19.........(1)
also (x+y)^2=x^2+2xy+y^2(i.e. 7^2)=49............(2)
subtract (1) from (2) leaving 3xy=30 or xy=10
x, y, z are 3 non zero positive integers such that x+y+z = 8 and xy+yz+zx = 20,
What would be minimum possible value of x*y^2*z^2
1/x + 1/y = 1/3
1/x + 1/z = 1/5
1/y + 1/z = 1/7
x/y = ??
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