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Question Number 106664 by deep last updated on 06/Aug/20

$$\mathrm{Factorise}:x^6+64y^6$$

Answered by som(math1967) last updated on 06/Aug/20

$${\left({\mathrm{x}}^2\right)}^3+{\left({4\mathrm{y}}^2\right)}^3$$$$=({\mathrm{x}}^2+{4\mathrm{y}}^2)({\mathrm{x}}^4-{\mathrm{x}}^2.{4\mathrm{y}}^2+{16\mathrm{y}}^4)$$$$=({\mathrm{x}}^2+{4\mathrm{y}}^2)({\mathrm{x}}^4-{4\mathrm{x}}^2{\mathrm{y}}^2+{16\mathrm{y}}^4)$$

Answered by nimnim last updated on 06/Aug/20

$${\left(x^2\right)}^3+{\left(4y^2\right)}^3$$$$(x^2+4y^2)\{{\left(x^2\right)}^2-\left(x^2\right)\left(4y^2\right)+{\left(4y^2\right)}^2\}$$$$(x^2+4y^2)(x^4-4x^2{y}^2+16y^4)$$

Answered by abdomathmax last updated on 07/Aug/20

$$\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^6+{64\mathrm{y}}^6\mathrm{roots}\mathrm{of}\mathrm{p}\left(\mathrm{x}\right)$$$$\mathrm{p}\left(\mathrm{x}\right)=0\iff {\mathrm{x}}^6=-{64\mathrm{y}}^6\Rightarrow {\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^6=-2^6\Rightarrow$$$${\left(\frac{\mathrm{x}}{2\mathrm{y}}\right)}^6=-1={\mathrm{e}}^{\mathrm{i}(2\mathrm{k}+1)\pi }\Rightarrow \frac{\mathrm{x}}{2\mathrm{y}}={\mathrm{e}}^{\frac{\mathrm{i}(2\mathrm{k}+1)\pi }{6}}\Rightarrow \mathrm{the}\mathrm{roots}\mathrm{are}$$$${\mathrm{x}}_{\mathrm{k}}=2\mathrm{y}{\mathrm{e}}^{\frac{\mathrm{i}(2\mathrm{k}+1)\pi }{6}}\mathrm{with}\mathrm{k}\in \left[[0,5]\right]\Rightarrow$$$$\mathrm{p}\left(\mathrm{x}\right)=\prod _{\mathrm{k}=0}^5(\mathrm{x}-{2\mathrm{ye}}^{\frac{\mathrm{i}(2\mathrm{k}+1)\pi }{6}})\left(\mathrm{factorisation}\mathrm{atC}\left(\mathrm{x}\right]\right)$$

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