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Question Number 5022 by love math last updated on 03/Apr/16

$$\frac{n^{1.2}-3\sqrt{3}}{n^{0.8}-\sqrt{3n^{0.8}}+3}-\sqrt{3}(\sqrt{3n^{0.8}}-1)$$$$$$$$Simplify$$

Commented by prakash jain last updated on 04/Apr/16

$$\mathrm{Substitute}\mathrm{to}\mathrm{ease}\mathrm{process}\mathrm{of}\mathrm{simplification}$$$$n^{.4}=a$$$$\sqrt{3}=b$$$$\frac{a^3-b^3}{a^2-ab+b^2}-b(ab-1)$$$$\mathrm{It}\mathrm{looks}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{probably}\mathrm{mistyped}$$$$\mathrm{the}\mathrm{first}\mathrm{probably}\mathrm{was}\frac{n^{1.2}-3\sqrt{3}}{n^{0.8}+\sqrt{3n^{0.8}}+3}$$$$\mathrm{In}\mathrm{that}\mathrm{case}\mathrm{first}\mathrm{part}\mathrm{simplifies}\mathrm{to}$$$$\frac{a^3-b^3}{a^2+ab+b^2}-b(ab-1)$$$$=\frac{(a-b)(a^2+ab+b^2)}{(a^2+ab+b^2)}-b(ab-1)$$$$=a-b-{ab}^2+b$$$$=a-{ab}^2$$$$=a(1-b^2)$$$$=-2n^{0.4}$$

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