If $a+b+c=6$ and $ab+bc+ca=11$, then the value of $a^3+b^3+c^3-3abc$ is:

The correct answer is 18.

Given:

$a+b+c=6$

$ab+bc+ca=11$

Formula:

We know the algebraic identity for the sum of cubes of three variables:

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca\left)\right)$

We can also rewrite $a^2+b^2+c^2$ using the identity ${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$.

Rearranging this gives:

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

Substituting this into the main algebraic identity, we get:

$a^3+b^3+c^3-3abc=(a+b+c)[{(a+b+c)}^2-3(ab+bc+ca\left)\right]$

Step-by-step Calculation:

Now, substitute the given values into the formula:

$a^3+b^3+c^3-3abc=6\times [6^2-3(11\left)\right]$

$a^3+b^3+c^3-3abc=6\times [36-33]$

$a^3+b^3+c^3-3abc=6\times 3=18$

Thus, the value of $a^3+b^3+c^3-3abc$ is 18.