Question : What is the value of $(a^2+b^2+4c^2-ab-2bc-2ca)(a+b+2c)$?
Option 1: $a^3 + b^3 + 8c^3 – 8abc$
Option 2: $a^3 + b^3 + 6c^3 – 4abc$
Option 3: $4a^3 + b^3 + 8c^3 – 9abc$
Option 4: $a^3 + b^3 + 8c^3 – 6abc$
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Correct Answer: $a^3 + b^3 + 8c^3 – 6abc$
Solution :
$(a^2+b^2+4c^2-ab-2bc-2ca)(a+b+2c)$
$= a^3+b^2a+4c^2a-a^2b-2abc-2ca^2 + a^2b+b^3+4c^2b-ab^2-2b^2c-2abc + 2a^2c+2b^2c+8c^3-2abc-4bc^2-4c^2a$
$= a^3+b^3+8c^3-2abc-2abc-2abc+b^2a-ab^2+4c^2a-4c^2a-a^2b+ a^2b-2ca^2 + 2a^2c-2b^2c+2b^2c-4bc^2+4c^2b$
$= a^3+b^3+8c^3-6abc$
Hence, the correct answer is $ a^3+b^3+8c^3-6abc$.
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