Explanation
Solution
Step 1: Analyze the Equation
The given equation is:
We can rewrite this as:
Using the identity A3+B3=(A+B)(A2−AB+B2), the roots satisfy:
(x−1+2)((x−1)2−2(x−1)+4)=0
The roots a,b,c are the roots of (x−1)3=−8.
Taking the cube root of both sides:
x−1=−2,−2ω,−2ω2
So, the roots are:
a=1−2=−1
b=1−2ω
c=1−2ω2
Step 2: Use the Property of the Determinant
The given determinant Δ is a standard form. It is known that:
2bc−a2c2b2amp;c2amp;2ac−b2amp;a2amp;b2amp;a2amp;2ab−c2=(a3+b3+c3−3abc)2
This is derived from the product of two circulant determinants involving a,b, and c.
Step 3: Calculate the Symmetric Sums of Roots
From the original equation x3−3x2+3x+7=0:
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Sum of roots: s1=a+b+c=−(1−3)=3
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Sum of roots taken two at a time: s2=ab+bc+ca=13=3
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Product of roots: s3=abc=−17=−7
Step 4: Evaluate a3+b3+c3−3abc
We use the algebraic identity:
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−(ab+bc+ca))
First, find a2+b2+c2:
a2+b2+c2=(a+b+c)2−2(ab+bc+ca)
Now, substitute these into the identity:
Step 5: Final Value of the Determinant
Final Answer:
The value of the determinant is 0. (Option D)
