If a + b + c = 6 then find the value of (a - 1) ^ 3 + (b - 2) ^ 3 + (c - 3) ^ 3 * i :

Options

3(a - 1)(b - 2)(c - 3)

3abc

(a - 1)(b - 2)(c - 3)

None

Given that \( a + b + c = 6 \), we need to find the value of the expression:

\[
(a - 1)^3 + (b - 2)^3 + (c - 3)^3 \times i
\]

First, recognize that the expression has the form of a well-known identity. The identity for the sum of cubes is:

\[
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
\]

If \( x + y + z = 0 \), the identity simplifies to:

\[
x^3 + y^3 + z^3 = 3xyz
\]

Now, let's assign:

\[
x = a - 1, \quad y = b - 2, \quad z = c - 3
\]

Notice that:

\[
x + y + z = (a - 1) + (b - 2) + (c - 3) = (a + b + c) - 6 = 6 - 6 = 0
\]

Since \( x + y + z = 0 \), the identity simplifies to:

\[
(a - 1)^3 + (b - 2)^3 + (c - 3)^3 = 3(a - 1)(b - 2)(c - 3)
\]

Hence, the value of the expression is:

\[
\boxed{3(a - 1)(b - 2)(c - 3)}
\]

Would you like further details on this solution? Here are five related questions you might find useful:

1. How does the identity for the sum of cubes change if \( x + y + z \neq 0 \)?
2. Can the same method be applied if the sum \( a + b + c \) was a different value?
3. How can we prove the identity \( x^3 + y^3 + z^3 = 3xyz \) when \( x + y + z = 0 \)?
4. What happens if we have complex numbers involved in the sum \( x + y + z \)?
5. How does the presence of \( i \) (imaginary unit) affect the calculation of cubes in algebraic expressions?

**Tip:** Always look for patterns and identities when dealing with algebraic expressions involving powers. This can significantly simplify your calculations.

Learn how to solve the problem involving the sum of cubes and complex numbers with a detailed step-by-step solution. This problem covers key algebraic concepts, including the identity for sum of cubes and its application to complex numbers. Suitable for high school students in Grades 10-12.

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