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# Is This A New Formula For Infinitely Nested Radicals?

$x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2\sqrt[3]{4a(x^2+3(4a)+(4a)^2)+x^2\dotsm}}}$ $$Derivation:$$I made use of the following identity; $$x+a=\sqrt[3]{(x+a)^3}\\=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}$$ In the 2nd step I wrote $$3ax^2$$ as $$2ax^2+ax^2$$.In the 3rd step I rearranged the terms.In the 4th step I factored out the $$a$$ from the first 3 terms and I factored out the $$x^2$$ from the last 2 terms.Manipulating $$x+2a=\sqrt[3]{x^3+3(2a)x^2+3(2a)^2x+(2a)^3}$$ in the same way,I got $$x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}$$.In the same way I got the expression for $$x+4a,x+8a,x+16a\dotsm$$.Now I observed that each of these can be substituted in the previous expression.Substituting my way upto $$x+a$$,I got the expression written at the beginning.Now,I want the people on Brilliant to verify this expression and to see whether it works.Please help me!!!

Note by Abdur Rehman Zahid
2 years, 2 months ago

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Put a = 0

R.H.S becomes 0 but L.H.S is still 'x'

So I don't think it's correct · 2 years, 2 months ago

Because you are dealing with infinite sums/roots, you have to be really careful of what the proper interpretation is.

Of course, it could be that the statement is not true for $$a = 0$$, but is true for all other real numbers. Staff · 2 years, 2 months ago

Just looking back at this note and thinking......That was a long time ago · 1 year ago

Putting $$a \,=\, 0$$ yields another infinitely nested series of $$x$$, which on simplification is equal to $$x$$. · 9 months, 2 weeks ago

Yup · 9 months, 2 weeks ago

Please explain how R.H.S becomes zero · 2 years, 2 months ago

Every term on R.H.S is multiple of 'a' you can see itself · 2 years, 2 months ago

If we rearrange the formula for $$x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}$$, you get $$x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}$$ In this expression if you replace a by 2a, you get $$x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}$$Replacing this expression in the expression for x +a, you get $$x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a))}}$$In this way if we continue deriving the expression for x+4a,x+8a,x+16a etc and replacing them in the original formula for x+a we will get the original formula .So R.H.S does not become zero you just have to replace the value of x+2a instead of the next radical and then set a=0 and after simplifying you get $$x=\sqrt[3]{x^3}$$ · 2 years, 2 months ago

Ah yes. Isn't it amazing how much can be achieved in a year? Staff · 1 year ago

Yeah.I didn't even know a thing back then.Brilliant has helped me a lot · 1 year ago