How many ways can you find to factorize \((ay+bx)^{3}+(ax+by)^{3}-(a^{3}+b^{3})(x^{3}+y^{3})\)?

**Hint**: The Solution is \(3abxy(a+b)(x+y)\)

I spent about half an hour getting it solved, obviously the way I worked wasn't efficient. I just used no more than the equality \(x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})\) and simplified it step by step.

Looking forward to your more brilliant solution. Thanks.

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## Comments

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TopNewest\( (ay+bx)^{3}+(ax+by)^{3}-(a^{3}+b^{3})(x^{3}+y^{3}) \)

\( = (ay+bx)^{3}+(ax+by)^{3}- (ax)^3 - (ay)^3 - (bx)^3 - (by)^3 \)

\( = (ay + bx)^3 + (-ay)^3 + (-bx)^3 + (ax + by)^3 + (-ax)^3 + (-by)^3 \)

Using the fact that if \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \), we have,

\( = 3(ay + bx)(-ay)(-bx) + 3(ax + by)(-ax)(-by) \)

\( = 3abxy(ay + bx + ax + by) \)

\( = 3abxy(a+b)(x+y) \)

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For cases where you strongly believe that there is a nice factorization, the approach that I like to use is the (extended) remainder factor theorem.

For example, let the expression be \( f(a, b, x, y) \). Observe that since \( f(0, b, x, y) = 0 \), this strongly suggests that \( a \mid f (a, b, x, y ) \). Similarly for the rest of the variables.

Next, since \( f( a, b, x, -x) = 0 \), this strongly suggests that \( (x+y) \mid f(a, b, x, y ) \). Similarly for \(a\) and \(b\).

Hence, this tells us that \( abxy (a+b)(x+y) \mid f(a,b,x,y) \). Since \( f(a,b,x,y) \) is a degree 6 polynomial, we get that \( f( a, b, x, y) = N abxy (a+b)(x+y) \). Substituting in values, we get that \( N = 3 \).

Note: This may not always be the case. The factor theorem in multiple variables doesn't always work out so nicely (hence the

strongly suggests), and the generalizations are in the field of algebraic geometry. However, looking for zeros of your expression is often a good way to get started guessing at what the factorization is.Log in to reply

Thanks, upvoted already :D

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