a+b+c - 6 , ab + bc+ ca - 11 Find the value of a3+b3 + c3

You have two given equations involving \(a\), \(b\), and \(c\):

1. \(a + b + c = 6\)
2. \(ab + bc + ca = 11\)

We need to find the value of \(a^3 + b^3 + c^3\).

To proceed, we can use the identity for the sum of cubes of three variables:

\[
a^3 + b^3 + c^3 - 3abc = (a + b + c)((a + b + c)^2 - 3(ab + bc + ca))
\]

Let's break it down step by step:

### Step 1: Calculate \((a + b + c)^2\)
\[
(a + b + c)^2 = 6^2 = 36
\]

### Step 2: Compute \((a + b + c)^2 - 3(ab + bc + ca)\)
\[
36 - 3 \times 11 = 36 - 33 = 3
\]

### Step 3: Use the identity for \(a^3 + b^3 + c^3\)
\[
a^3 + b^3 + c^3 - 3abc = (a + b + c) \times 3 = 6 \times 3 = 18
\]
Thus, 
\[
a^3 + b^3 + c^3 = 18 + 3abc
\]

To find the exact value of \(a^3 + b^3 + c^3\), we need to know the value of \(abc\). Without this, we can only express the result in terms of \(abc\).

So the final expression is:
\[
a^3 + b^3 + c^3 = 18 + 3abc
\]

Would you like help finding \(abc\) or any other detail?

### Related questions:
1. How do we calculate \(abc\) from these equations?
2. Can you derive the identity for \(a^3 + b^3 + c^3\)?
3. What happens if \(a = b = c\)?
4. How would the result change if the constants in the original equation were different?
5. Can this method be applied to find higher powers of \(a + b + c\)?

**Tip**: Memorize the identity for the sum of cubes when dealing with symmetric expressions—it saves a lot of time!

a + b + c = 6, ab + bc + ca = 11. Find the value of a^3 + b^3 + c^3.

Find the value of a^3 + b^3 + c^3 using the sum of cubes identity. Given a + b + c = 6 and ab + bc + ca = 11, learn step-by-step how to solve this problem using algebraic identities. Ideal for high school students (Grades 10-12).

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