Question

Show that\( a^3 + b^3 + c^3 - 3abc = \dfrac{1}{2}(a+b+c)\left\{(a-b)^2 + (b-c)^2 + (c-a)^2\right\}\)

Hint:

Remember the identity: \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\). Then use \((a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)\).

Answer 1

Step 1: Start with the right-hand side.}
We take the expression:
\[ \dfrac{1}{2}(a+b+c)\left\{(a-b)^2 + (b-c)^2 + (c-a)^2\right\} \]
Step 2: Expand the terms inside the bracket.}
Now,
\[ (a-b)^2 = a^2 - 2ab + b^2 \] \[ (b-c)^2 = b^2 - 2bc + c^2 \] \[ (c-a)^2 = c^2 - 2ca + a^2 \] Adding these, we get:
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 \] \[ = a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 \] \[ = 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca \] \[ = 2(a^2 + b^2 + c^2 - ab - bc - ca) \]
Step 3: Substitute this back into the expression.}
Substituting in the right-hand side, we get:
\[ \dfrac{1}{2}(a+b+c)\cdot 2(a^2 + b^2 + c^2 - ab - bc - ca) \] \[ = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) \]
Step 4: Use the standard identity.}
We know the algebraic identity:
\[ (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc \]
Step 5: Conclude the proof.}
Therefore,
\[ \dfrac{1}{2}(a+b+c)\left\{(a-b)^2 + (b-c)^2 + (c-a)^2\right\} = a^3 + b^3 + c^3 - 3abc \] Hence, proved that:
\[ \boxed{a^3 + b^3 + c^3 - 3abc = \dfrac{1}{2}(a+b+c)\left\{(a-b)^2 + (b-c)^2 + (c-a)^2\right\}} \]