My symbolic calculator was able to factor $$6(8^x+27^x)-7(12^x+18^x),$$ resulting in $(3^x+2^x)(2 \cdot 3^x-3 \cdot 2^x)(3 \cdot 2^x - 2 \cdot 3^x).$ I could not get the factorization by hand, and could only verify it by expanding the factored expression. I am asking if anyone knows how to approach such factorizations, even if only a route for this example. Any information is appreciated. [note: the problem of finding the real zeroes appeared in a facebook math group: https://www.facebook.com/groups/319322575482767/permalink/1359233248158356]
2026-04-24 19:40:33.1777059633
Factoring exponential expressions
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The point is that $8^x = (2^x)^3$, $27^x = (3^x)^3$, $12^x = 3^x \cdot (2^{x})^2$ and $18^x = 2^x (3^x)^2$. If we let $s = 2^x$ and $t = 3^x$, the expression to be factored becomes $$ 6 (s^3 + t^3) - 7 (s^2 t + s t^2)$$ Now can you do it?