Calculating $a^3 + b^3 + c^3$ Using Algebraic Identities

This problem requires us to find the value of $a^3 + b^3 + c^3$ given the values of $a+b+c$, $ab+bc+ca$, and $abc$. We can solve this using standard algebraic identities.

Given Information

We are provided with the following values:

• $a + b + c = 7$
• $ab + bc + ca = 11$
• $abc = -1$

Relevant Algebraic Identity

The key identity that relates $a^3 + b^3 + c^3$ to the given sums and products is:

$\hspace{4em} a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$

We can rearrange this identity to solve for $a^3 + b^3 + c^3$:

$\hspace{4em} a^3 + b^3 + c^3 = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca)) + 3abc$

Finding $a^2+b^2+c^2$

Before we can use the identity for $a^3 + b^3 + c^3$, we need to find the value of $a^2+b^2+c^2$. We know the identity:

$\hspace{4em} (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$

We can rearrange this to find $a^2+b^2+c^2$:

$\hspace{4em} a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)$

Substitute the given values $a+b+c = 7$ and $ab+bc+ca = 11$:

$\hspace{4em} a^2+b^2+c^2 = (7)^2 - 2(11)$

$\hspace{4em} a^2+b^2+c^2 = 49 - 22$

$\hspace{4em} a^2+b^2+c^2 = 27$

Calculating $a^3+b^3+c^3$

Now we have all the necessary values to use the identity for $a^3 + b^3 + c^3$:

$\hspace{4em} a^3 + b^3 + c^3 = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca)) + 3abc$

Substitute the values we know:

• $a+b+c = 7$
• $a^2+b^2+c^2 = 27$
• $ab+bc+ca = 11$
• $abc = -1$

Plugging these into the formula:

$\hspace{4em} a^3 + b^3 + c^3 = (7)(27 - 11) + 3(-1)$

First, calculate the term inside the parenthesis:

$\hspace{4em} 27 - 11 = 16$

Now substitute this back:

$\hspace{4em} a^3 + b^3 + c^3 = (7)(16) + 3(-1)$

Perform the multiplications:

$\hspace{4em} (7)(16) = 112$

$\hspace{4em} 3(-1) = -3$

Finally, combine the results:

$\hspace{4em} a^3 + b^3 + c^3 = 112 - 3$

$\hspace{4em} a^3 + b^3 + c^3 = 109$

Conclusion

The value of $a^3 + b^3 + c^3$ is 109.

Revision Table: Algebraic Identities Used

Additional Information: Symmetric Polynomials

The expressions like $a+b+c$, $ab+bc+ca$, and $abc$ are examples of elementary symmetric polynomials in three variables $a, b,$ and $c$.

• The first elementary symmetric polynomial is $e_1 = a+b+c$. This is the sum of the variables.
• The second elementary symmetric polynomial is $e_2 = ab+bc+ca$. This is the sum of products of the variables taken two at a time.
• The third elementary symmetric polynomial is $e_3 = abc$. This is the product of the variables taken three at a time.

Any symmetric polynomial in $a, b, c$ (a polynomial that remains unchanged if any two variables are swapped) can be expressed in terms of these elementary symmetric polynomials $e_1, e_2, e_3$. The expression $a^3+b^3+c^3$ is a symmetric polynomial, and its relationship with $e_1, e_2, e_3$ is given by the identity we used:

$\hspace{4em} a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))$

Or, expressed using $e_1, e_2, e_3$:

$\hspace{4em} a^3 + b^3 + c^3 - 3e_3 = e_1((a+b+c)^2 - 2(ab+bc+ca) - e_2)$

$\hspace{4em} a^3 + b^3 + c^3 - 3e_3 = e_1(e_1^2 - 2e_2 - e_2)$

$\hspace{4em} a^3 + b^3 + c^3 = e_1^3 - 3e_1e_2 + 3e_3$

Let's quickly check if this derived formula gives the same result using the given values $e_1=7, e_2=11, e_3=-1$:

$\hspace{4em} a^3 + b^3 + c^3 = (7)^3 - 3(7)(11) + 3(-1)$

$\hspace{4em} a^3 + b^3 + c^3 = 343 - 231 - 3$

$\hspace{4em} a^3 + b^3 + c^3 = 112 - 3$

$\hspace{4em} a^3 + b^3 + c^3 = 109$

Both methods yield the same result, confirming the correctness of the calculations and the identities used.