I wanna know if $$\left[(x^2+h)^3-(x+h)\right]^{1/2}-[x^3-x]^{1/2}=\left[(x+h)^3-(x+h)-x^3+x\right]^{1/2}.$$
$\endgroup$ 52 Answers
$\begingroup$Does $a^2 - b^2 = (a-b)^2\;$?
No, Because $$(a-b)^2 = a^2 -2ab +b^2 \neq a^2 - b^2$$ Alternatively $$a^{1/2} - b^{1/2} = \sqrt a -\sqrt b \neq \sqrt{a-b}$$ Simply take $a = 4, b = 9$. Then we have $2 - 3 = -1$. Whereas $\sqrt{2-3}$ is undefined in the reals, and which will never be equal to a negative real number.
$\endgroup$ $\begingroup$after squaring two times and factorizing we obtain $$x \left( x-1 \right) \left( {x}^{10}+{x}^{9}+6\,{x}^{8}h+{x}^{8}+6\,{ x}^{7}h+15\,{x}^{6}{h}^{2}-{x}^{7}+9\,{x}^{5}{h}^{2}+18\,{x}^{4}{h}^{3 }-{x}^{6}-6\,{x}^{5}h-9\,{x}^{4}{h}^{2}+9\,{x}^{2}{h}^{4}-{x}^{5}-6\,{ x}^{4}h-15\,{x}^{3}{h}^{2}-18\,{x}^{2}{h}^{3}-9\,x{h}^{4}-12\,{x}^{3}h -12\,{x}^{2}{h}^{2}-4\,x{h}^{3}-12\,{x}^{2}h-12\,x{h}^{2}-4\,{h}^{3}+4 \,xh+4\,h \right) =0$$ and now?
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