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If x + y = 7 & x^3 + y^3 = 133 Then xy=??

Where x,y both are real numbers.

x is 5 y is 2 so xy=10

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

If x, y, z are 3 non zero positive integers such that x+y+z = 8 and xy+yz+zx = 20, then What would be minimum possible value of x*y^2*z^2

If 1/x + 1/y = 1/3 1/x + 1/z = 1/5 1/y + 1/z = 1/7 then x/y = ??