Question : If $(a+b+c) \neq 0$, then $(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$ is equal to:
Option 1: $a^3+b^3-c^3-3abc$
Option 2: $a^3-b^3+c^3-3abc$
Option 3: $a^3+b^3+c^3-3abc$
Option 4: $a^3+b^3+c^3+3abc$
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Correct Answer: $a^3+b^3+c^3-3abc$
Solution :
Given: $(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$
Simplifying this expression, we have:
$(a^3+ab^2+ac^2–a^2b–abc–ca^2+a^2b+b^3+bc^2–ab^2–b^2c–abc+a^2c+b^2c+c^3–abc–bc^2–c^2a)$
= $(a^3+b^3+c^3–3abc)$
Hence, the correct answer is $(a^3+b^3+c^3–3abc)$.
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