If (a + b + c) = 16, and (a² + b² + c²) = 90, find the value of (ab + bc + ca).

The correct answer is 83.

We are given two pieces of information:

(a + b + c) = 16
(a² + b² + c²) = 90

We need to find the value of (ab + bc + ca).

Step 1: Recall the key algebraic identity.

The most important identity here is the square of a trinomial:

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

This identity is the foundation of our solution. It directly links the three quantities we care about.

Step 2: Substitute the known values into the identity.

We know (a + b + c) = 16, so its square is:

{\left(16\right)}^2=90+2(ab+bc+ca)

256=90+2(ab+bc+ca)

Step 3: Isolate the term (ab + bc + ca).

Subtract 90 from both sides:

256-90=2(ab+bc+ca)

166=2(ab+bc+ca)

Now divide both sides by 2:

ab+bc+ca=\frac{166}{2}=83

Result: The value of (ab + bc + ca) = 83.