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factor 16r^6-2p^6
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16(r-p)(r^2+rp+p^2)(r^3+p^3)??
\[2((2r ^{2})^{3}-((p)^{2})^{3}\]
so us formula
\[2(a ^{3}-b ^{3})\]
Almost. The answer is actually \[16r^6-2p^6=2(8r^6-p^6)=2(2r^3-p^3)(2r^3+p^3)\] If you want, you can further simplify \[2r^3-p^3 \text{ and } 2r^3+p^3\] using the fomula \[x^3+y^3=(x+y)(x^2-xy+y^2)\] and \[x^3-y^3=(x-y)(x^2+xy+y^2)\]
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