Sure, let's solve the problem step-by-step.
Given:
1. [tex]\( a + b + c = 5 \)[/tex]
2. [tex]\( ab + bc + ca = 10 \)[/tex]
We want to find [tex]\( a^3 + b^3 + c^3 - 3abc \)[/tex].
To approach this problem, we use a known algebraic identity:
[tex]\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)\left((a + b + c)^2 - 3(ab + bc + ca)\right) \][/tex]
Plugging in the given expressions for [tex]\( a + b + c \)[/tex] and [tex]\( ab + bc + ca \)[/tex], we can simplify this identity.
First, let's calculate:
[tex]\[ (a + b + c)^2 = 5^2 = 25 \][/tex]
Next:
[tex]\[ 3(ab + bc + ca) = 3 \times 10 = 30 \][/tex]
Now, substitute these values into our identity:
[tex]\[ a^3 + b^3 + c^3 - 3abc = (5)\left(25 - 30\right) \][/tex]
Simplify inside the parentheses:
[tex]\[ 25 - 30 = -5 \][/tex]
Thus, we have:
[tex]\[ a^3 + b^3 + c^3 - 3abc = 5 \times (-5) = -25 \][/tex]
So, we have shown that given [tex]\( a + b + c = 5 \)[/tex] and [tex]\( ab + bc + ca = 10 \)[/tex], it follows that:
[tex]\[ a^3 + b^3 + c^3 - 3abc = -25 \][/tex]
Therefore, the proof is complete.
