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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
Values are 4,2,2
So answer is 64
Solving we got
X*y^2*z^2 = X(x^2-8x+20)
Since X has a positive value and we have to get minimum then putting X=1
So 169 is the answer
ans is 64, for minimum value, x,y& z will be 4,2 &2,
4, 2, 2
2, 3+sqr(5), 3-sqr(5)
The answer is 32
x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be
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