Question

should the answer of 81m⁴-(n-m)⁴ after factorisation using identities be (10m²+n²-2mn)(2m+n)(4m-n)?

Answer 1

$$\begin{gathered} 81{\mathrm{m}}^4-{\left(\mathrm{n}-\mathrm{m}\right)}^4 \\ ={\left(9{\mathrm{m}}^2\right)}^2-{\left\{{\left(\mathrm{n}-\mathrm{m}\right)}^2\right\}}^2 \\ =\left[9{\mathrm{m}}^2-{\left(\mathrm{n}-\mathrm{m}\right)}^2\right]\left[9{\mathrm{m}}^2+{\left(\mathrm{n}-\mathrm{m}\right)}^2\right] \\ =\left[{\left(3\mathrm{m}\right)}^2-{\left(\mathrm{n}-\mathrm{m}\right)}^2\right]\left[9{\mathrm{m}}^2+{\mathrm{n}}^2+{\mathrm{m}}^2-2\mathrm{mn}\right] \\ =\left[3\mathrm{m}+\left(\mathrm{n}-\mathrm{m}\right)\right]\left[3\mathrm{m}-\left(\mathrm{n}-\mathrm{m}\right)\right]\left(10{\mathrm{m}}^2+{\mathrm{n}}^2-2\mathrm{mn}\right) \\ =\left(2\mathrm{m}+\mathrm{n}\right)\left(4\mathrm{m}-\mathrm{n}\right)\left(10{\mathrm{m}}^2+{\mathrm{n}}^2-2\mathrm{mn}\right) \end{gathered}$$