x×(y+z)=x×y+x×z
x=-2/3,y=-4/6,z=-7/9
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Form a quadratic equation whose Roots are x,y,z
Now, let the equation be ax3+bx2+cx+d=0
So, Sum x+y+z=−b/a
xy+yz+xz=c/a
xyz=−d/a
If a=1 , equation will be x3−6x2+15x−14=0
So, x=2 is a root of equation : the equation can be written as (x−2)(x2−4x+7)=0
Since x2−4x+7=0 has imaginary roots which are 2+isqrt3 and 2−isqrt3
So, the only real solution for this equation is x=2
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