Question

Factorise. 2root2Xcube-3root3Ycube

Answer 1

$2\sqrt{2}\ x^3-3\sqrt{3}\ y^3 \\ \\ \\ x^3\sqrt{8}-y^3\sqrt{27} \\ \\ \\ (x\sqrt{2})^3-(y\sqrt{3})^3 \\ \\ \\ (x\sqrt{2}-y\sqrt{3})((x\sqrt{2})^2+(x\sqrt{2} \cdot y\sqrt{3})+(y\sqrt{3})^2) \\ \\ \\ (x\sqrt{2}-y\sqrt{3})(2x^2+xy\sqrt{6}+3x^2) \\ \\ \\ \\ \\ \\ \therefore\ 2\sqrt{2}\ x^3-3\sqrt{3}\ y^3=\bold{(x\sqrt{2}-y\sqrt{3})(2x^2+xy\sqrt{6}+3x^2)}$

 

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