Simplify the expression 2(sin^6θ + cos^6θ) − 3(sin^4θ + cos^4θ) + 1 and determine its constant value for all real θ.

Introduction / Context:
This problem is another example of simplifying trigonometric expressions involving higher powers of sine and cosine. The expression 2(sin^6θ + cos^6θ) − 3(sin^4θ + cos^4θ) + 1 may look complicated, but with the standard identity sin^2θ + cos^2θ = 1 and some algebraic identities for sums of powers, it reduces to a constant independent of θ.

Given Data / Assumptions:

• Expression: 2(sin^6θ + cos^6θ) − 3(sin^4θ + cos^4θ) + 1.
• θ is any real angle.
• Basic identity: sin^2θ + cos^2θ = 1.

Concept / Approach:
As in similar problems, we set s = sin^2θ and c = cos^2θ. We then express sin^6θ + cos^6θ and sin^4θ + cos^4θ in terms of s and c, and further in terms of the product sc. After substitution, the expression becomes a linear combination of 1 and sc, and the sc terms cancel out, leaving a simple constant value.

Step-by-Step Solution:
Let s = sin^2θ and c = cos^2θ. Then s + c = 1. Compute sin^6θ + cos^6θ = s^3 + c^3. Use a^3 + b^3 = (a + b)^3 − 3ab(a + b). Thus s^3 + c^3 = (s + c)^3 − 3sc(s + c) = 1^3 − 3sc·1 = 1 − 3sc. Next compute sin^4θ + cos^4θ = s^2 + c^2. Use a^2 + b^2 = (a + b)^2 − 2ab. Thus s^2 + c^2 = (s + c)^2 − 2sc = 1^2 − 2sc = 1 − 2sc. Now substitute into the original expression: 2(sin^6θ + cos^6θ) − 3(sin^4θ + cos^4θ) + 1 = 2(1 − 3sc) − 3(1 − 2sc) + 1. Expand each part: 2 − 6sc − 3 + 6sc + 1. Combine like terms: (2 − 3 + 1) + (−6sc + 6sc) = 0 + 0 = 0.
Verification / Alternative check:
Check with a few values of θ. For θ = 0°, sin θ = 0 and cos θ = 1, so the expression becomes 2·1 − 3·1 + 1 = 0. For θ = 45°, sin θ = cos θ = √2/2, so sin^2θ = cos^2θ = 1/2 and both sin^6θ and cos^6θ equal 1/8. Substitution again yields zero, confirming the identity.

Why Other Options Are Wrong:
Values −1, 1 or 2 would not hold for all θ. For example, at θ = 0° the expression equal to 0 contradicts these options. The value −2 is also impossible because the expression can be shown to be zero for many different angles.

Common Pitfalls:
A common error is misinterpreting sin^6θ as sin 6θ. Another pitfall is misusing the algebraic identities for sums of squares and cubes, leading to extra terms or incorrect coefficients. By carefully substituting s = sin^2θ, c = cos^2θ and using s + c = 1, the computation becomes straightforward.

Final Answer:
The simplified value of the expression is 0 for all real θ.