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# How do I solve this?

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Factor \$(ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3.\$

Jun 3, 2023

#1
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We can factor this expression as follows:

(ab + ac + bc)^3 - a^3b^3 - a^3c^3 - b^3c^3 = (ab + ac + bc)(ab + ac + bc)^2 - a^3b^3 - a^3c^3 - b^3c^3 = (ab + ac + bc)(ab^2 + ac^2 + bc^2 + 2abc) - a^3b^3 - a^3c^3 - b^3c^3 = (ab + ac + bc)(a^2b^2 + a^2c^2 + b^2c^2 + 2abc + a^2bc + a^2cb + b^2ac + c^2ab) - a^3b^3 - a^3c^3 - b^3c^3 = (ab + ac + bc)(a^2(b^2 + c^2) + b^2(c^2 + a^2) + c^2(a^2 + b^2) + 2(abc + a^2bc + a^2cb + b^2ac)) - a^3b^3 - a^3c^3 - b^3c^3 = (ab + ac + bc)(a^2(b + c)^2 + b^2(a + c)^2 + c^2(a + b)^2 + 2(abc + a^2bc + a^2cb + b^2ac)) - a^3b^3 - a^3c^3 - b^3c^3 = (ab + ac + bc)(a^2(b + c)^2 + b^2(a + c)^2 + c^2(a + b)^2 + 2abc(a + b + c)) - (a^3b + a^3c + b^3a + b^3c + c^3a + c^3b) = (ab + ac + bc)(a^2(b + c)^2 + b^2(a + c)^2 + c^2(a + b)^2 + 2abc(a + b + c) - (a^2b + a^2c + b^2a + b^2c + c^2a + c^2b)) = (ab + ac + bc)(a^2(b + c)^2 + b^2(a + c)^2 + c^2(a + b)^2 + 2abc(a + b + c) - (a^2 + ab + ac + b^2 + bc + c^2)) = (ab + ac + bc)(a^2(b + c)^2 + b^2(a + c)^2 + c^2(a + b)^2 + 2abc(a + b + c) - (a + b + c)(a^2 + b^2 + c^2)).

The expression in the parentheses is a perfect square, so we can factor it as follows:

(ab + ac + bc)(a^2(b + c)^2 + b^2(a + c)^2 + c^2(a + b)^2 + 2abc(a + b + c) - (a + b + c)(a^2 + b^2 + c^2)) = (ab + ac + bc)(a + b + c)((a + b)^2 + (b + c)^2 + (c + a)^2 - (a^2 + b^2 + c^2)) = (ab + ac + bc)(a + b + c)(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc).

Therefore, the complete factorization is:

(ab + ac + bc)(a + b + c)(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc).

Jun 3, 2023