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In this multi-part problem, we will consider this system of simultaneous equations:
3x+5y-6z =2,
5xy-10yz-6xz = -41,
xyz=6.
Let a=3x, b=5y, and c=-6z.
Determine the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c.
Make sure to enter your answer in terms of t and only t, in expanded form.
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3x + 5y - 6z = 2, (1)
5xy - 10yz - 6xz = -41, (2)
xyz = 6 (3)
Now multiply the equation (2) by 3 (both sides), and multiply the equation (3) by -3*5*6 = -90 (both sides).
You will get an equivalent system
3x + 5y - 6z = 2, (1')
15xy - 30yz - 18xz = -41*3 (2')
-90xyz = 6*(-90) (3')
It is the same as
3x + 5y - 6z = 2, (1'')
(3x)*(5y) + (5y)*(-6z) + (3x)*(-6z) = -123 (2'')
(3x)*(5y)*(-6z) = -540 (3'')
Using variables a, b and c, you can rewrite the last system equivalently as
a + b + c = 2, (1''')
ab + bc + ac = - 123, (2''')
abc = -540. (3''')
Now the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c IS
t^3 - 2t^2 + (-123)t - (-540) = 0 ( <<<---+++ as the Vieta's theorem states (!) )
or, which is the same,
t^3 - 2t^2 - 123t + 540 = 0.
Solved.